## NCERT Solutions for Class 11 Chemistry Chapter 2: Structure of Atom

2.1. (i) Calculate the number of electrons which will together weigh one gram.

(ii) Calculate the mass and charge of one mole of electrons.

Ans. (i) Mass of one electron = 9.10939 × 10–31 kg

Number of electrons that weigh 9.10939 × 10–31 kg = 1

Number of electrons that will weigh 1 g = (1 × 10–3 kg)

$$=\frac{1}{9.10939×10^{\normalsize -31}}×1×10^{\normalsize -3}$$

= 0.1098 × 10–3+31

= 0.1098 × 1028

= 1.098 × 1027

(ii) 1 mole = 6.022 × 1023 atoms

Mass of one electron = 9.10939 × 10–31 kg

Mass of one mole of electron = (6.022 × 1023) × (9.10939 ×10–31 kg) = 5.48 × 10–7 kg

Charge on one electron = 1.6022 × 10–19 coulomb

Charge on one mole of electron = (1.6022 × 10–19 C) × (6.022 × 1023) = 9.65 × 104 coulomb

2.2. (i) Calculate the total number of electrons present in one mole of methane.

(ii) Find (a) the total number, and (b) the total mass of neutrons in 7 mg of 14C. (Assume that mass of a neutron = 1.675 × 10–27 kg).

(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of NH 3 at STP. Will the answer change if the temperature and pressure are changed?

Ans. (i) One mole of methane (CH4) has molecules

= 6.022 × 1023

Number of electrons present in 1 molecule of methane (CH4) = (1 × (6) + 4 × (1)) = 10

Number of electrons present in 1 mole i.e., 6.023 × 1023 molecules of methane
= 6.022×1023 × 10

= 6.022×1024 electrons

(ii) (a) Number of atoms of 14C in 1 mole

= 6.023 × 1023

Since 1 atom of 14C contains

= 14 – 6 = 8 neutrons,

Therefore, the number of neutrons in 14 g of 14C = (6.023 × 1023) × 8

Now,

Since, 14 g of 14C contains (6.022 × 1023  × 8) neutrons.

Number of neutrons in 7 mg
= (6.022 × 1023× 8 × 7) / 14000 mg

= 2.4092 × 1021

(b) Mass of one neutron = 1.675 × 10–27 kg

Mass of total neutrons in 7 mg of 14C

= (2.4092 × 1021) (1.675 × 10–27 kg)

= 4.0352 × 10–6 kg

(iii) (a) 1 mole of NH3 = (1 × (14) + 3 × (1)) g of NH3 = 17 g of NH3

= 6.022 × 1023 molecules of NH3

Total number of protons present in 1 molecule of NH3

= (1 × (7) + 3 × (1)) = 10

Number of protons in 6.023 × 1023 molecules of NH3

Now,

= (6.023 × 1023) × (10)

= 6.023 × 1024

Now,

17 g of NH3 contains (6.023 × 1024) protons.

Number of protons in 34 mg of NH3

= (6.023 × 1024) × 34 mg/17000 mg

= 1.2046 × 1022

(b) Given that,

Mass of one proton = 1.675 × 10–27 kg

Therefore, total mass
of protons in 34 mg of NH3

= (1.675 × 10–27 kg) (1.2046 × 1022)

= 2.0176 × 10–5 kg

No, the answer will not change if the temperature and pressure are changed because the number of protons, electrons, and neutrons in an atom are independent of temperature and pressure conditions.

2.3. How many neutrons and protons are there in the following nuclei?

$$^{13}\text{C}_6, ^{16}\text{O}_8, ^{24}\text{Mg}_{12}\space^{56}_{26}\space\text{Fe},\space^{88}_{38}\text{Sr}$$

Ans. For 13C6

Given: Mass number (A) = 13

Atomic number (Z) of carbon = 6

Atomic number = Number of protons (p) = 6

Number of neutrons (n)

= (mass number) – (Atomic number)

= 13 – 6 = 7

For 16O8

Given: Mass number (A) of Oxygen = 16

Atomic number (Z) of Oxygen = 8

Atomic number = number of protons (p) = 8

Number of neutrons (n)

= (Mass number) – (Atomic number)

= 16 – 8 = 8

For 24Mg12

Given: Mass number (A) of magnesium = 24

Atomic number (Z) of magnesium = 12

Atomic number = Number of protons (p) = 12

Number of neutrons (n)
= (Mass number) – (Atomic number)

= 24 – 12 = 12

$$\text{For}\space^{56}_{26}\space\text{Fe}$$
Given: Mass number (A) of iron = 56

Atomic number (Z) of iron = 26

Atomic number = Number of protons (p) = 26

Number of neutrons (n)

= (Mass number) – (Atomic number)

= 56 – 26

= 30

$$\text{For}\space^{88}_{38}\space\text{Sr}$$

Given: Mass number (A) strontium = 88

Atomic number (Z) of strontium = 38

Atomic number = Number of protons (p) = 38

Number of neutrons (n)

= (Mass number) – (Atomic number)

= 88 – 38

= 50

2.4. Write the complete symbol for the atom with the given atomic number (Z) and Atomic mass (A)

(i) Z = 17, A = 35

(ii) Z = 92, A = 233

(iii) Z = 4, A = 9

$$\textbf{Ans}.\space\text{(i)}\space ^{35}\text{Cl}_{17}\\\text{(ii)}\space^{233}\text{U}_{92}\\\text{(iii)}\space^{9}\text{Be}_{4}$$

2.5. Yellow light emitted from a sodium lamp has a wavelength (l) of 580 nm. Calculate the frequency (n) and wave number $$(\bar{v})$$ of the yellow light.

Ans. We know that, the relation between wavelength and frequency is given by,

$$\lambda=\frac{c}{v}\\\text{or}\space\text{v}=\frac{c}{\lambda}\\\text{where, n = Frequency of light}\\\text{c = Velocity of light in vacuum = 3 × 10}^{8} \text{m/s}\\\lambda=\text{Wavelength of yellow light = 580 nm}=580 × 10^{\normalsize–9} \text{m (Given)}\\\text{Substituting the values in the above formula, we get,}\\\text{v}=\frac{3×10^{8}\space\text{m/s}}{580×10^{\normalsize-9}\space\text{m}}\\= 5.17 × 10^{14} s^{–1}\\\text{Now, wave number of yellow light is given as,}\\\bar v=\frac{1}{\lambda}\\=\frac{1}{580×10^{-9}}\text{m}\\= 1.724 × 10^{6} \text{m}^{–1}$$

2.6. Find energy of each of the photons which

(i) correspond to light of frequency 3 × 1015 Hz.

(ii) have wavelength of 0.50 Å.

Ans. (i) Energy (E) of a photon is given by the expression,

E = hv

Where, h = Planck’s constant = 6.626 × 10–34 Js

n = frequency of light = 3 × 1015 Hz

= 3 × 1015 s–1 Given)

Substituting the values we get,

E = (6.626 × 10–34 Js) × (3 × 1015 s–1)

E = 1.988 × 10–18 J

(ii) Energy (E) of a photon having wavelength (n) is given by the expression,

$$\text{E}=\frac{hc}{\lambda}\\\text{Where}, \\\text{h = Planck’s constant = 6.626 × 10}^{\normalsize–34} \text{Js}\\\text{c = Velocity of light in vacuum =}3×10^{8}\text{m/s}\\\lambda=\text{Wavelength = 0.50 Å or 0.50 x 10}^{\normalsize –10} \text{m}\\\text{Substituting values in expression we get,}\\= (6.626 × 10^{–34} \text{Js}) ×\frac{(3×10^{8}\space\text{ms}^{-1})}{(0.5×10^{-10}\text{m})}\\= 3.98 × 10^{–15} \text{J}$$

2.7. Calculate the wavelength, frequency and wave number of a light wave whose period is 2.0 × 10–10 s.

$$\textbf{Ans.}\space\text{Frequency (n) of light}=\frac{1}{\text{period}}\\= 1 / 2.0 × 10^{\normalsize–10} \text{s}.\\= 5.0 × 10^{9} s^{–1}\\\text{Wavelength of light = c/v}\\=\frac{3×10^{8}\text{ms}^{\normalsize-1}}{5.0×10^{9}s^{\normalsize-1}}\\= 6 × 10^{–2} \text{m}\\\text{Wave}\space \text{number}(\bar v)=\frac{1}{\text{wavelength} (\lambda)}\\=\frac{1}{6×10^{\normalsize-2}}\text{m}\\= 1.66 × 10^{1} \text{m}^{–1} = 16.66 \space\text{m}$$

2.8. What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?

Ans. Given:

λ = wavelength of light = 4000 pm

= 4000 ×10–12 m

c = velocity of light in vacuum = 3 × 108 m/s

h = Planck’s constant = 6.626 × 10–34 Js

Energy of 1 photon = hv $$=\frac{hc}{\lambda}=\frac{(6.626×10^{-34}\space\text{Js})(3×10^{8}\space\text{ms}^{-1})}{(4×10^{-9} \space\text{m})}\\\space\text{Number of photons that provide 1 J energy}\\=\frac{1\text{J}}{\text{Energy of one photon in joule}}\\=\frac{(1\space\text{J})(4×10^{-9}\text{m})}{(6.626×10^{-34}\text{Js})(3×10^{8}\text{ms}^{-1})}\\= 2.012 × 10^{16} \text{photons}.$$ Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are 2.012 × 1016.

2.9. A photon of wavelength 4 × 10–7 m strikes on metal surface, the work function of the metal being 2.13 eV. Calculate

(i) the energy of the photon (eV),

(ii) the kinetic energy of the emission, and

(iii) the velocity of the photoelectron (1 eV = 1.6020 × 10–19 J).

$$\textbf{Ans.}\space\text{(i)}\space\text{Energy (E) of a photon}=\frac{\text{hc}}{\lambda}\\\text{Where, h = Planck’s constant} = 6.626 × 10^{\normalsize –34} \text{Js}\\\text{c = velocity of light in vacuum}=3×10^{8}\space\text{m/s}\\\lambda=\text{wavelength of photon}=4×10^{\normalsize-7}\text{m}\\\text{Substituting the values, we get,}\\\text{E}=\frac{(6.626×10^{\normalsize -34})(3×10^{8})}{(4×10^{\normalsize -7})}\\=4.965×10^{\normalsize -19}\text{J}\\\text{Hence, the energy of the photon is} 4.97 × 10^{\normalsize –19} \text{J}.\\\text{Now, the energy of photon in eV can be calculated as :}\\=\frac{1\text{eV}}{(1.602×10^{\normalsize -19}\text{J})}×(4.97×10^{\normalsize -19}\text{J})\\=3.1\text{eV}$$

(ii) Kinetic energy of the emission

= E – Work function

i.e., Kinetic energy of emitted electrons

= (3.1 – 2.13)

= 0.97 eV

Hence, the kinetic energy of emission is 0.97 eV.

(iii) The velocity of a photoelectron (v) can be calculated by the expression,

$$\frac{1}{2}\text{mv}^{2}=\text{hv – hv}_0\\\Rarr\space\text{v}=\sqrt{\frac{2(\text{hv-hv}_0)}{\text{m}}}\\\text{where} (hv – hv_0)\space\text{is kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron.}\\\text{Substituting the values in the given expression of v:}\\\text{v}=\sqrt{\frac{2×(0.97×1.6020×10^{\normalsize-19}\text{J})}{9.10939×10^{-31}\text{kg}}}\\=\sqrt{0.3418×10^{12}\text{m}^{2}\text{s}^{-2}}\\\text{v}=5.84×10^{5}\space\text{ms}^{-1}\\\text{Hence, the velocity of the photoelectron is}\space5.84 × 10^{5}\space\text{ms}^{\normalsize –1}.$$

2.10. Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol–1.

Ans. Given: l = 242 nm = 242 × 10–9 m,

c = 3 × 108 ms–1, h = 6.626 × 10–34 Js

$$\text{E}=\frac{\text{hc}}{\lambda}\\\therefore\text{E}=\frac{(6.626×10^{-34}\text{Js})(3×10^{8}\text{ms}^{-1})}{(242×10^{-9}\space\text{m})}\\= 0.0821 × 10^{–17} \text{J}\\\text{Ionisation energy per mol (E)}=\frac{(0.0821×10^{-17})(6.022×10^{23}\space\text{mol}^{-1}\text{J})}{1000}\\= 494\space\text{kJ}\space\text{mol}^{-1}$$

2.11. A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57μm. Calculate the rate of emission of quanta per second.

Ans. Energy of one photon,

$$\text{E = hv}=\frac{\text{hC}}{\lambda}\\\text{Where,}\\\text{c = velocity of light in vacuum} = 3 × 10^{8} \text{m/s}\\\text{h = Planck’s constant =}\space6.626 × 10^{\normalsize –34}\text{Js}\\\lambda=0.57\space\mu\text{m}=0.57×10^{\normalsize -6} \text{m} (\text{Given})\\\text{Substituting the values, we get,}\\\text{E}=\frac{(6.626×10^{\normalsize -34})(3×10^{8})}{0.57×10^{\normalsize -6}}\\= 34.87 × 10^{\normalsize –20} \text{J}\\\text{Rate of emission of quanta per second}\\=\frac{25}{34.87×10^{\normalsize -20}}\\= 7.169 × 10^{19} \text{s}^{\normalsize –1}$$

2.12. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (v0 ) and work function (W0) of the metal.

Ans. Threshold wavelength of radiation (λ0)

= 6800 × 10–10 m or 6.8 × 10–7

$$\text{Threshold frequency of the metal}\space(\text{v}_0)=\frac{c}{\lambda_0}\\=\frac{3×10^{8}\space\text{ms}^{\normalsize -1}}{6.8×10^{\normalsize -7}}\\= 4.41 × 10^{14} \text{s}^{\normalsize –1}$$

Now,

Work function (W0) of the metal = hv0

= (6.626 × 10–34 Js) (4.41 × 1014 s–1)

= 2.922 × 10–19 J

2.13. What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n = 2?

Ans. The transition from ni = 4 to an energy level with nf = 2, will give rise to a spectral line of the Balmer series. The energy involved in the transition is given by the relation,

$$\text{E}=2.18×10^{\normalsize-18}\bigg[\frac{1}{n_i^2}-\frac{1}{n_f^2}\bigg]\\\text{Solving the equation we get,}\\\Delta\text{E}=2.18×10^{\normalsize-18}\bigg[\bigg(\frac{1}{4^{2}}\bigg)-\bigg(\frac{1}{2^{2}}\bigg)\bigg]\\2.18×10^{-18}\bigg[\bigg(\frac{1-4}{16}\bigg)\bigg]\\=2.18×10^{\normalsize-18}\bigg[\frac{-3}{16}\bigg]\\\text{or},\space\Delta \text{E}=-4.0875×10^{\normalsize -19}\text{J}\\\text{The negative sign indicates that the energy is emitted during the transition.}\\\text{Now, the wavelength of transmission is given by the expression,}\\\lambda=\frac{hc}{\text{E}}$$

where, c = velocity of light in vacuum

= 3 × 108 m/s

h = Planck’s constant = 6.626 × 10–34 Js

Substituting the values we get, $$=\frac{(6.626×10^{\normalsize-34})(3×10^{8})}{(-4.0875×10^{\normalsize-19})}$$

= 4.86 × 10–7 m

or, λ = 486 nm

Hence, the wavelength of light emitted is 486 nm.

2.14. How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n = 1 orbit).

Ans. The expression of energy is given by

$$\text{E}_n=-\frac{(2.18×10^{\normalsize-18})Z^{2}}{n^{2}}\\\text{Where,}\\ \text{Z = atomic number of the atom}\\ \text{n = principal quantum number For ionization from n}_1 = 5 to n = \text{infinity}\\\Delta\text{E}=\text{E}_\infty-\text{E}_5\\=-\frac{(2.18×10^{\normalsize-18})1^{2}}{\infty^{2}}-\frac{(-2.18×10^{\normalsize-18})1^{2}}{5^{2}}\\-\frac{(-2.18×10^{-18})1^{2}}{5^{2}}\\=0.0872 × 10^{-18} \text{J}\\\text{or},\space\Delta \text{E}=8.72×10^{-20}\text{J}\\\text{Now},\\\text{Energy required for n}_1 = 1 \text{to n} = ∞\\\Delta \text{E}=\text{E}_\infty-\text{E}_1\\=-\frac{(2.18×10^{-18})1^{2}}{\infty^{2}}-\frac{(-2.18×10^{-18})1^{2}}{1^{2}}$$

= 2.18 × 10 –18 J

Thus, according to the above calculation, less energy is required to ionize an electron in the 5th orbital of hydrogen atom as compared to that in the ground state.

2.15. What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the ground state?

Ans. Following transitions are possible, when the excited electron of an H atom in n = 6 drops to the ground state,

Therefore, a total number of 15 lines will be obtained in the emission spectrum i.e., (5 + 4 + 3 + 2 + 1).

The number of spectral lines produced when an electron in the nth level drops down to the ground state is given by

$$\frac{n(n-1)}{2}\\\text{Number of spectral lines =}\frac{6(6-1)}{2}(n=6)\\=\frac{6×5}{2}=15$$

2.16. (i) The energy associated with the first orbit in the hydrogen atom is –2.18 × 10–18 J atom–1. What is the energy associated with the fifth orbit?

(ii) Calculate the radius of Bohr’s fifth orbit for hydrogen atom.

Ans. (i) Energy associated with the fifth orbit of hydrogen atom can be calculated as:$$\text{E}_n=\frac{(2.18×10^{-18})Z^{2}}{n^{2}}\\\text{For H, n = 5 and Z = 1}\\\text{E}_5=-\frac{(2.18×10^{-18})1^{2}}{5^{2}}\\\text{E}_5=-8.72 × 10^{-20}\text{J}\\\text{(ii) The formula to calculate the radius of Bohr’s nth orbit for hydrogen atom is given by,}\\\ _n=\frac{52.9(n^{2})}{Z}\text{pm}\\\text{Now, to calculate the radius of Bohr’s fifth orbit for hydrogen atom,}\\\text{put n = 5}\\\text{Z = 1 (atomic number of hydrogen)}\\r_{g}=\frac{52.9×10^{-12}×5^{2}\text{m}}{1}$$

= 1.326 × 10–9 m

= 1.3225 nm.

Hence, the radius of Bohr’s fifth orbit for hydrogen atomis 1.3225 nm.

2.17. Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Ans. For the Balmer series, ni = 2.

Wavenumber is given by,

$$=\bigg[\frac{1}{(2)^{2}}-\frac{1}{(n_f)^{2}}\bigg](1.097×10^{7}\space \text{m}^{-1})\\\text{As wave number is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, wave number has to be the smallest.}\\\text{For wave number to be minimum, n}_f \text{should be minimum. Thus, taking nf = 3 , we get,}\\\text{wave number =}\bigg[\frac{1}{(2)^{2}}-\frac{1}{(3)^{2}}\bigg](1.097×10^{7}\space\text{m}^{-1})\\=\bigg[\frac{1}{4}-\frac{1}{9}\bigg](1.097×10^{7} \text{m}^{-1})\\=\bigg[\frac{5}{36}\bigg](1.097×10^{7}\space\text{m}^{-1})\\=1.523×10^{6}\space\text{m}^{\normalsize-1}$$

2.18. What is the energy in joules, required to shift the electron of the hydrogen atom from the first Bohr orbit to the fifth Bohr orbit and what is the wavelength of the light emitted when the electron returns to the ground state? The ground state electron energy is –2.18 × 10–11 ergs.

Ans Energy (E) of the nth Bohr orbit of an atom is given by, $$\text{E}_n=-\frac{(2.18×10^{\normalsize-18})Z^{2}}{n^{2}}\\\text{Where},\\\text{Z = atomic number of the atom}\\\text{n = principal quantum number}\\\text{Ground state energy = –2.18 ×10}^{\normalsize-11}\text{ergs}(\text{Given})\\=-2.18 × 10^{\normalsize-11}×10^{\normalsize-7}\text{J}(1\space\text{J = 10}^{7}\text{ergs})\\=-2.18 × 10^{\normalsize-18}\text{J}\\\text{Energy required to shift the electron from n = 1 to n = 5 is given as:}\\\text{E = E}_5 – \text{E}_1\\\text{E}_n=-\frac{(2.18×10^{\normalsize-18})1^{2}}{5^{2}}-(-2.18×10^{\normalsize-18})\\=(2.18×10^{\normalsize-18})\bigg(1-\frac{1}{25}\bigg)\\=(2.18×10^{\normalsize-18})\bigg(\frac{24}{25}\bigg)\\2.098×10^{\normalsize-18}\text{J}$$

$$\text{wavelength of emitted light =}\frac{hc}{\text{E}}\\=\frac{(6.626×10^{\normalsize-34})(3×10^{8})}{2.098×10^{-18}}\\=9.498×10^{\normalsize -8}\space\text{m}$$

2.19. The electron energy in hydrogen atom is given by

$$\text{E}_n=\frac{(-2.18×10^{\normalsize-18})}{n^{2}}\text{J}.$$ Calculate the energy required to remove an electron completely from the n = 2 orbit. What is the longest wavelength of light in cm that can be used to cause this transition ?

Ans. Given that the electron energy in hydrogen atom $$\text{E}_n=\frac{(-2.18×10^{\normalsize-18})}{n^{2}}\text{J}$$

The energy required is the difference in the energy when the electron jumps from orbit with n = ∞ to orbit with n = 2.

The eneregy required

$$\Delta\text{E}=\text{E}_\infty-\text{E}_2$$ $$=\bigg[\frac{(-2.18×10^{\normalsize-18})}{\infty^{2}}\bigg]-\bigg[\frac{(-2.18×10^{\normalsize-18})}{2^{2}}\bigg]\\0-\bigg[\frac{(-2.18×10^{\normalsize-18})}{4}\bigg]\text{J}$$

= 0.545 × 10–18 J

ΔE = 5.45 × 10-19

Now, the longest wavelength of light in cm used to cause the transition can be calculated by the formula,

$$\lambda=\frac{hc}{\text{E}}\\=\frac{(6.626×10^{\normalsize-34})×(3×10^{8})}{5.45×10^{\normalsize-19}}\\=3647×10^{\normalsize-10}\text{m}\space\text{or}\space 3.645×10^{\normalsize-7}\text{m}$$

2.20. Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 ms–1.

Ans. According to de Broglie’s equation,

$$\lambda=\frac{h}{mv}\\\text{Where,}\\\lambda \text{= wavelength of moving particle}\\\text{m = mass of particle =}9.109 × 10^{-31}\text{kg}\\\text{v = velocity of particle = 2.05 × 10}^{7}\text{ms}^{-1}(\text{Given})\\\text{h = Planck’s constant =}\space 6.626 × 10^{-34} \\\text{Substituting the values in the expression of}\space\lambda,\\\lambda=\frac{(6.626×10^{\normalsize -34})}{(9.09×10^{\normalsize-31})×(2.05×10^{7}\text{ms}^{-1})}\\= 3.54 × 10^{\normalsize-11}\text{m}\\\text{Hence, the wavelength of the electron is}\\= 3.54 × 10^{\normalsize-11}\text{m}.$$

2.21. The mass of an electron is 9.1 × 10-31 kg. If its K.E. is 3.0 × 10–25 J, calculate its wavelength.

Ans. According to de Broglie’s equation,

$$\lambda=\frac{h}{mv}\\\text{Where}, \lambda = \text{wavelength of moving particle}\\\text{m}=\text{mass of particle}=9.1×10^{\normalsize-31}\text{kg}\\\text{v = velocity of particle}\\ \text{h} = \text{Planck’s constant} = 6.626 × 10^{\normalsize–34}\\\text{Now calculating value of v,}\\\text{We know that K.E =}\frac{1}{2}\text{mv}^{2}\\v=\sqrt{\frac{2\text{K.E}}{m}}\\=\sqrt{\frac{2(3.0×10^{\normalsize-25})}{(9.1×10^{\normalsize-31})\text{kg}}}$$

v = 811. 579 ms-1

$$\lambda=\frac{h}{mv}(\text{de-Broglie’s equation})\\=\frac{6.626×10^{-34}}{(9.1×10^{-31})(811.579)}\\=8.96×10^{\normalsize-7}\text{m}$$

Thus, the wavelength of electron is 8.96 × 10–7 m or 8967 Å.

2.22. Which of the following are isoelectronic species i.e., those having the same number of electrons? Na+, K+, Mg2+, Ca2+, S2–, Ar

Ans. An isoelectronic series is a group of ions that all have the same number of electrons. Therefore, first of all,calculate the number of electrons for each ions. A cation is formed by loss of electrons and an anion is formed by gain of electrons.

Atomic number of Na = 11, K = 19, Mg = 12, Ca = 20, S = 16, Ar = 18

Number of electrons in (Na+) = (11 – 1) = 10
Number of electrons in K+ = (19 – 1) = 18 Number of electrons in Mg2+= (12 – 2) = 10
Number of electrons in Ca2+ = (20 – 2 ) = 18
Number of electrons in sulphur S2– = (16 +2) = 18
Number of electrons in argon (Ar) = 18
Hence, isoelectronic species are:
Na+ and Mg2+ (10 electrons each)
K+, Ca2+, S2– and Ar (18 electrons each)

2.23. (i) Write the electronic configurations of the following ions:

(a) H
(b) Na+
(c) O2–
(d) F

(ii) What are the atomic numbers of elements whose outermost electrons are represented by

(a) 3s1
(b) 2p3
(c) 3p5?

(iii) Which atoms are indicated by the following configurations?

(a) [He] 2s1
(b) [Ne] 3s2 3p3
(c) [Ar] 4s2 3d1.

Ans. (i) (a) Electronic configuration of H = 1s2

(b) Electronic configuration of Na+ = 1s2 2s2 2p6

(c) Electronic configuration of O2– ion = 1s2 2s2 2p6

(d) Electron configuration of F ion = 1s2 2s2 2p6

(ii) (a) The electronic configuration of the given element will be 1s2 2s2 2p6 3s1

Thus, the number of electrons present in the atom of the element = 2 + 2 + 6 + 1
= 11

Therefore, atomic number of the given element is 11, i.e., Na

(b) The electronic configuration of given element will be 1s2 2s2 2p3

Thus, the number of electrons present in the atom of the element = 2 + 2 + 3 = 7

Therefore, atomic number of the given element is 7 i.e., N.

(c) The electronic configuration of the given element will be 1s2 2s2 2p6 3s2 3p6 4s2 3d6

Thus, the number of electrons present in the atom of the element = 2 + 2 + 6 + 2 + 6 + 2 + 6 = 26

Therefore, atomic number of the given element is 26, i.e., Fe

(iii) (a) [He] 2s1 = Atomic number of He = 2 thus, 2 + 1 = 3 i.e., Lithium (Li)

(b) [Ne] 3s2 3p3 = Atomic number of Ne = 10 thus, 10 + 2 + 3 = 15 i.e., Phosphorous (P)

(c) [Ar] 4s2 3d1 = Atomic number of Ar = 18 thus, 18 + 2 + 1 = 21 i.e., Scandium (Sc).

2.24. What is the lowest value of n that allows g orbitals to exist?

Ans. For g-orbitals, l = 4.

The value for l for Principal quantum number n is given by (n – 1).

∴ For l = 4, lowest value of n = 5.

2.25. An electron is in one of the 3d-orbitals. Give the possible values of n, l and m1 for this electron.

Ans. For the 3d-orbital:

Principal quantum number (n) = 3

Azimuthal quantum number (l) = (n – 1) = 2

Magnetic quantum number (ml) = (2l +1) = 5

= – 2, – 1, 0, 1, 2

2.26. An atom of an element contains 29 electrons and 35 neutrons. Deduce (i) the number of protons and (ii) the electronic configuration of the element.

Ans. (i) In a neutral atom the number of protons is equal to the number of electrons.

∴ Number of protons in the atom of the given element = 29

(ii) The electronic configuration of the atom is 1s2 2s2 2p6 3s2 3p63d104s1

2.27. Give the number of electrons in the species H2+, H2 and O2+.

Ans. For H2+

Number of electrons present in dihydrogen cation (H2+) is = 2 – 1 = 1

For H2

Number of electrons present in dihydrogen molecule (H2) = 2

For O2+

Number of electrons present in dioxygen cation (O2+) = 16 -1 = 15

2.28. (i) An atomic orbital has n = 3. What are the possible values of l and ml ?

(ii) List the quantum numbers (ml and l) of electrons for 3d-orbital.

(iii) Which of the following orbitals are possible? 1p, 2s, 2p and 3f.

Ans. (i) n = 3 (Given)

For a given value of n, l can have values from 0 to (n – 1). Hence, For n = 3 l = 0, 1, 2

For a given value of l, ml can have (2l + 1) values.

Thus,

When l = 0, ml = 0.

When l = 1, ml = –1, 0, +1

When l = 2, m = –2, –1, 0, +1, +2

(ii) For 3d-orbital, l = 2.

For a given value of l, ml can have (2l + 1) values i.e., 5 values.

For l = 2 ml = –2, – 1, 0, 1, 2

(iii) Among the given orbitals only 2s and 2p are possible. 1p and 3f cannot exist.

1p is not possible because when n = 1, l = 0 only (for p, l = 1)

2s is possible because when n = 2, l = 0, 1 (for s, l = 0)

2p is possible because when n = 2, l = 0, 1 (for p, l = 1)

3f is not possible because when n = 3, l = 0, 1, 2 (for f, l = 3).

2.29. Using s, p, d notations, describe the orbital with the following quantum numbers.

(a) n = 1, l = 0;

(b) n = 3; l = 1

(c) n = 4; l = 2;

(d) n = 4; l = 3.

Ans. (a) For n = 1, l = 0, The orbital is 1s.

(b) For n = 3 and l = 1, the orbital is 3p.

(c) For n = 4 and l = 2 The orbital is 4d.

(d) For n = 4 and l = 3 The orbital is 4f.

2.30. Explain, giving reasons, which of the following sets of quantum numbers are not possible.

$$(a) n = 0, l = 0, m_l =0\space\space m_s=+\frac{1}{2}\\(b) n = 1, l = 0, m_l = 0\space\space m_s=-\frac{1}{2}\\\text{(c) n = 1, l = 1, m}_l = 0\space\space m_s=+\frac{1}{2}\\\text{(d) n = 2, l = 1, m}_l = 0\space\space m_s=-\frac{1}{2}\\\text{(e) n = 3, l = 3, m}_l = –3\space\space m_s=+\frac{1}{2}\\(f) n = 3, l = 1, ml = 0,\space\space m_s=+\frac{1}{2}$$

Ans. (a) The given set of quantum numbers is not possible because the value of the principal quantum number (n) cannot be zero.

(b) The given set of quantum numbers is possible.

(c) The given set of quantum numbers is not possible because for a given value of n, ‘l’ can have values from zero to (n – 1). Thus, for n = 1, l = 0 and not 1.

(d) The given set of quantum numbers is possible.

(e) The given set of quantum numbers is not possible because, for n = 3, l = 0 to (3 – 1). Thus, l = 0 to 2 i.e., 0, 1, 2. Therefore, for n = 3, l is not equal to +3.

(f)  The given set of quantum numbers is possible.

2.31. How many electrons in an atom may have the following quantum numbers?

$$\textbf{(a) n = 4, m}_\textbf{s} =\text{-}\frac{\text{1}}{\text{2}}\\\textbf{(b) n = 3, l = 0}$$

Ans. (a) Total number of electrons in an atom for a value of n is given by = 2n2

Therefore, for n = 4,

Total number of electrons = 2 (4)2 = 32 And half out of these electrons will have

$$m_s=-\frac{1}{2}.\text{Thus, the total electrons with}\\m_s\bigg(-\frac{1}{2}\bigg)\text{is 16 because according to Pauli’s exclusion principle two electrons in an orbital cannot have the same spin.}$$

(b) For n = 3, maximum electrons can be = 2n2
Total number of electrons = 18.

But n = 3, l = 0 indicates that the electrons are present in the 3s orbital. Therefore, the number of electrons having n = 3 and l = 0 is only 2.

2.32. Show that the circumference of the Bohr orbit for the hydrogen atom is an integral multiple of the de Broglie wavelength associated with the electron revolving around the orbit.

Ans. A hydrogen atom contains only one electron. According to Bohr’s postulate, the angular momentum of that electron is given by ,

$$mvr = \frac{hn}{2\neq}\space\space...(i)\\\text{where n} = 1,2,3 ……\\\text{According to de Broglie’s equation,}\\\lambda=\frac{h}{mv}\\\text{or}\space\space\text{mv}=\frac{h}{\lambda}\space\space...(ii)\\\text{Substituting the value of mv in equation (i), we get}\\\frac{hr}{\lambda}=\frac{n}{2\neq}\\\frac{r}{\lambda}=\frac{n}{2\neq}\\2\pi r= n\lambda\space\space...(iii)$$

where ‘2pr’ represents the circumference of the Bohr orbit (r),

Therefore from above equation (eq. (iii)), it is proved that the circumference of the Bohr orbit of the hydrogen atom is an integral multiple of de Broglie’s wavelength associated with the electron revolving around the orbit.

2.33. What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He+ spectrum?

Ans. The wave number associated with the Balmer transition, n = 4 to n = 2 of He+ spectrum is given by:

$$\vec{v}=\frac{1}{\lambda}=R_HZ^{2}\bigg(\frac{1}{n_1^{2}}-\frac{1}{n_2^{2}}\bigg)\\\text{Where,}\space n_1 = 2 \space\text{and}\space n_2 = 4\\\text{R}_H=\text{Rydberg’s constant}\\z=\text{atomic number of helium}\\\text{Substituting the values we get,}\\\frac{1}{\lambda}=R_H(2)^{2}\bigg(\frac{1}{2^{2}}-\frac{1}{4^{2}}\bigg)\\=4R_H\bigg(\frac{1}{4}-\frac{1}{16}\bigg)\\=4R_H\bigg(\frac{4-1}{16}\bigg)\\=\frac{3R_H}{4}\\\text{or},\space \lambda=\frac{4}{3R_H}$$

According to the question, the desired transition for hydrogen will have the same wavelength as that of He+.

$$\text{R}_\text{H}(Z)^{2}\bigg(\frac{1}{n_1^{2}}-\frac{1}{n_2^{2}}\bigg)=\frac{1}{\lambda}\\\text{R}_H(1)^{2}\bigg(\frac{1}{n_1^{2}}-\frac{1}{n_2^{2}}\bigg)=\frac{3R_H}{4}\\\bigg(\frac{1}{n_1^{2}}-\frac{1}{n_2^{2}}\bigg)=\frac{3}{4}\\n_1=1\space\text{and}\space n_2=2$$

∴ The transition for n2 = 2 to n1 = 1 in hydrogen spectrum will have the same wavelength as Balmer transition for n = 4 to n = 2 of He+ spectrum.

2.34. Calculate the energy required for the process:

$$\text{He}^+(\text{g})\xrightarrow{}\text{He}^{2+}(\text{g})+e^{-}$$

The ionization energy for the H atom in the ground state is 2.18 ×10–18 J atom–1.

Ans. Energy associated with hydrogen-like species is given by,

2.35. If the diameter of a carbon atom is 0.15 nm, calculate the number of carbon atoms which can be placed side by side in a straight line across length of scale of length 20 cm long.

Ans. We know that,

1 m = 100 cm

$$\text{1\space cm\space=}\space\frac{1}{100}=10^{\normalsize-2}$$

Length of the scale = 20 cm

= 20×10–2 m

∴ 1nm = 10–9m
Diameter of a carbon atom = 0.15 nm

= 0.15 × 10–9 m

One carbon atom occupies 0.15 × 10–9 m.

Number of carbon atoms that can be placed side by side in a straight line

$$=\frac{20×10^{-2}\text{m}}{0.15×10^{-9}\text{m}}\\= 1.33 × 10^9$$

2.36. 2 ×108 atoms of carbon are arranged side by side. Calculate the radius of carbon atom if the length of this arrangement is 2.4 cm.

Ans. Given:

Length of the given arrangement = 2.4 cm or 2.4 × 10–2 m

Number of carbon atoms present = 2 × 108

Therefore, diameter of carbon atom,

$$=\frac{2.4×10^{-2}}{2×10^{8}}\\= 1.2 × 10^{–10} \text{m}\\\text{Radius of each carbon atom =}\frac{\text{D}}{2}\\=\frac{1.2×10^{-10}}{2}\\=6.0 × 10^{–11} \text{m}$$

2.37. The diameter of zinc atom is 2.6 Å. Calculate (a) radius of zinc atom in pm and (b) number of atoms present in a length of 1.6 cm if the zinc atoms are arranged side by side lengthwise.

$$\textbf{Ans.}\space\text{(a) Radius of zinc atom =}\frac{\text{Diameter}}{2}\\\frac{2.6}{2}$$

= 1.3 Å

= 1.3 × 10–10 m

= 130 × 10–12 m

= 130 pm

(b) Length of the arrangement

= 1.6 cm = 1.6 × 10–2 m

Diameter of zinc atom

= 2.6 Å = 2.6 × 10–10 m

∴ Number of zinc atoms present along the

$$\text{length = }\frac{1.6×10^{2}\text{m}}{2.6×10^{-10}\text{m}} = 0.615 × 108 = 6.15 × 107 2.38. A certain particle carries 2.5 × 10–16C of static electric charge. Calculate the number of electrons present in it. Ans. Magnitude of charge (q) = 2.5 × 10–16 C Charge on one electron (e) = 1.602 × 10–19 C$$\text{Number of electrons present =}\frac{2.5×10^{-16}\text{C}}{(1.602×10^{-19}\text{C})}\$

= 1560 electrons

Hence, the number of electrons present are 1560 electrons.

2.39. In Milikan’s experiment, static electric charge on the oil drops has been obtained by shining X-rays. If the static electric charge on the oil drop is –1.282 × 10–18C, calculate the number of electrons present on it.

Ans. Charge on the oil drop = –1.282 ×10–18 C (Given)

Charge on one electron = –1.6022 × 10–19 C

Number of electrons present on the oil drop

$$=\frac{-1.282×10^{\normalsize-18}\text{C}}{1.6022×10^{\normalsize-19}\text{C}} = 8.0$$

Thus, the number of electrons present is 8.0.

2.40. In Rutherford’s experiment, generally the thin foil of heavy atoms, like gold, platinum etc. have been used to be bombarded by the a-particles. If the thin foil of light atoms like aluminium etc. is used, what difference would be observed from the above results?

Ans. As we know that Al is less malleable and ductile as compare to gold, the results would be different if aluminium is used. Also, due to less positive charge and low penetrating power of aluminium sheets, deflection would be not enough. Also, it will block the path of a particle as they have low penetrating power. Moreover, If light atoms are use,their nuclei will be light & moreover,they will have small positive charge on the nucleus. Hence, the number of particles deflected back & those deflect through some angle will be negligible.

2.41. Symbols 79Br35 or 79Br can be written, whereas symbols 35Br79 or 35Br are not acceptable. Answer briefly.

Ans. Atomic mass (A) and atomic number (Z) of and element can be represented as AXZ. Hence, Symbols 79Br35 can be accepted. Also 79Br can be written it shows atomic number and it is fixed. But mass number an element depends upon the relative abundance of its isotopes and can be different. Therefore, it is necessary to mention atomic mass.

2.42. An element with mass number 81 contains 31.7% more neutrons as compared to protons. Assign the atomic symbol.

Ans. Let the number of protons in the element be x.
Number of neutrons in the element

= x + 31.7% of x

= x + 0.317 x

= 1.317 x

Now, Mass number of the element = 81 (Given)

or, (Number of protons + number of neutrons) = (x + 1.317 x) = 81

2.317 x = 81

$$\text{x}=\frac{81}{2.317}$$

= 34.95 or 35

Thus, the number of protons is 35.

To assign atomic symbol, atomic number and mass number should be known.

Now, number of protons = atomic number = 35

Mass number = 81 (Given)

The element with atomic number (Z) 35 is bromine.

Therefore, the symbol will be 81Br35.

2.43. An ion with mass number 37 possesses one unit of negative charge. If the ion contains 11.1% more neutrons than the electrons, find the symbol of the ion.

Ans. Let the number of electrons in the ion carrying a negative charge be x. Then, Number of neutrons present = x + 11.1% of x

= x + 0.111 x

= 1.111 x

Number of electrons in the neutral atom = (x – 1)

(When an ion carries a negative charge, it carries an extra electron)

∴ Number of protons in the neutral atom = x – 1

Mass number of the ion = 37 (Given)

∴ (x – 1) + 1.111x = 37

2.111x = 38x

x = 18

Therefore no of protons = atomic number

= x – 1 = 18 – 1 = 17

∴ The symbol of the ion is 37Cl17.

2.44. An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol to this ion.

Ans. Let the number of electrons present in ion, E3+ = x

∴ Number of neutrons in it = x + 30.4% of x = 1.304 x

As the ion is tripositive,

Number of electrons in neutral atom = x + 3

∴ Number of protons in neutral atom = x + 3

Now mass number = no of protons + no of neutrons

56 = x + 3 + 1.304x

or 2.304x = 53

or x = 23

Therefore, atomic number of the ion (or element)

= x +3

= 23 + 3 = 26

The element with atomic number 26 is Iron (Fe) and the corresponding ion is Fe3+.

∴ The symbol of the ion 5626Fe3+.

2.45. Arrange the following type of radiations in increasing order of frequency: (a) radiation from microwave oven (b) amber light from traffic signal (c) radiation from FM radio (d) cosmic rays from outer space and (e) X-rays.

Ans. The increasing order of frequency is as follows:

Radiation from FM radio < amber light < radiation from microwave oven < X-rays < cosmic rays

The increasing order of wavelength is as follows:

Cosmic rays < X-rays < radiation from microwave ovens < amber light < radiation of FM radio

2.46. Nitrogen laser produces a radiation at a wavelength of 337.1 nm. If the number of photons emitted is 5.6 × 1024, calculate the power of this laser.

Ans. Power of laser = Energy with which it emits photons

P = E

$$\text{Power of the laser, E}=\frac{nhc}{\lambda}\\\text{Where,}\\\text{n = number of photons emitted} = 5.6 × 10^{24}\\\text{c = velocity of light in vacuum =}3 × 10^{8} \text{m/s}\\\text{h = Planck’s constant} = 6.626 × 10^{–34} \text{Js}\\\lambda=\text{wavelength of radiation = 337.1 nm or 337.1×10}^{-9}\text{m}\\\text{Substituting the values in the given expression of Energy (E):}\\\text{E}=\frac{(5.6×10^{24})(3×10^{8}\text{ms}^{\normalsize-1})(6.626×10^{\normalsize-34}\text{Js})}{(600×10^{\normalsize-9}\text{m})}$$

= 0.3302 × 107 J

= 3.33 × 106J

Hence, the power of the laser is 3.33 × 106J.

2.47. Neon gas is generally used in the sign boards. If it emits strongly at 616 nm, calculate (a) the frequency of emission, (b) distance traveled by this radiation in 30 s (c) energy of quantum and (d) number of quanta present if it produces 2 J of energy.

$$\textbf{Ans.}\text{(a) Frequency of emission}\space v=\frac{c}{\lambda}$$

where c = velocity of light in vacuum

= 3 × 108 m/s

λ = wavelength of radiation = 616 nm

= 616 × 10–9 m

$$\text{Therefore,}\space v=\frac{3×10^{8}\text{ms}^{-1}}{616×10^{-9}\text{m}}\\= 0.00487 × 10^{8} × 10^{9} \space \text{s}^{\normalsize–1}$$

or, v = 4.87 × 1014 s–1

(b) Velocity of radiation, (c) = 3.0 × 108 m s–1

Distance travelled by this radiation in 30 s

= velocity × time

= (3.0 × 108 ms–1) (30 s)

= 9.0 × 109 m

(c) Energy of quantum (E) = hv

= (6.626 × 10–34 Js) (4.87 × 1014 s–1)

Energy of quantum (E) = 32.27 × 10–20 J

(d)  Energy of one photon (quantum)

= 32.27 × 10–20 J.

Therefore, 32.27 × 10–20 J of energy is present in 1 quantum. Number of quanta in

$$\text{2J}\space\text{of energy}=\frac{2\text{J}}{32.27×10^{-20}\text{J}}\\=6.19×10^{18}$$

2.48. In astronomical observations, signals observed from the distant stars are generally weak. If the photon detector receives a total of 3.15 × 10–18 J from the radiations of 600 nm, calculate the number of photons received by the detector.

Ans. We know that,

$$\text{E}=\frac{hc}{\lambda}$$

Where,

c = velocity of light in vacuum = 3 × 108m/s

h = Planck’s constant = 6.626 × 10–34 Js

= 600 nm = 600 × 10–9 m

Substituting the values in the given expression of E, we get,

$$\text{E}=\frac{(3×10^{8}\text{ms}^{\normalsize-1})(6.626×10^{\normalsize-34}\text{Js})}{(600×10^{\normalsize-9}\text{m})}$$

= 3.313 × 10–19 J

Now,

Number of photons received with 3.15 × 10–18 J

$$\text{energy}=\frac{3.15×10^{\normalsize-18}}{3.313×10^{\normalsize-19}}$$

= 9.5 or 10 (approx.)

(since number of photons cannot be fraction).

2.49. Lifetimes of the molecules in the excited states are often measured by using pulsed radiation source of duration nearly in the nano second range. If the radiation source has the duration of 2 ns and the number of photons emitted during the pulse source is 2.5 × 1015, calculate the energy of the source.

Ans. Frequency of radiation v is given by

$$\text{v}=\frac{1}{2×10^{-9}}\space(2 \text{ns} = 2 × 10^{\normalsize –9}\text{s})$$

Now,

Energy (E) of source = Nhn

Where,

N = number of photons emitted (2.5 × 1015)

(Given)

h = Planck’s constant = (6.626 × 10–34)

v = frequency of radiation = 5.0 × 108 s–1

Substituting the values in the given expression of (E), we get

E = (2.5 × 1015) (6.626 × 10–34 Js) (5.0 × 108 s–1)

= 8.282 × 10–10 J

Hence, the energy of the source (E) is 8.282 × 10–10 J.

2.50. The longest wavelength doublet absorption transition is observed at 589 and 589.6 nm. Calculate the frequency of each transition and energy difference between two excited states.

Ans. For first transition

$$\text{Frequency (v}_{1})=\frac{c}{\lambda_1}\\=\frac{3×10^{8}}{589×10^{\normalsize-9}}\\= 5.093 × 10^{14} \text{s}^{\normalsize–1}$$

$$\text{Frequency (v}_{1})=\frac{c}{\lambda_1}\\=\frac{3×10^{8}}{589×10^{\normalsize-9}}\\= 5.093 × 10^{14} \text{s}^{\normalsize–1}\\\text{Frequency(v}_2)=\frac{c}{\lambda_2}\\=\frac{3×10^{8}}{589.6×10^{\normalsize-9}}$$

= 5.088 × 1014 s–1

Energy difference (ΔE) between excited states

= E1 – E2

= (hv1– hv2)

= h(v1 – v2)

= (6.626 × 10–34) (5.093 × 1014 – 5.088 × 1014)

= (6.626 × 10–34) (0.005 × 1014)

= 0.0331 × 10–20

= 3.31 × 10–22 J

2.51. The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

Ans. (a) Threshold wavelength

Work function (W0) for caesium atom = 1.9 eV or 1.9 × 1.602 × 10–19 J (1 eV= 1.602 × 10–19 J)

Now,

$$\text{W}_0=\frac{hc}{\lambda_0}\\\lambda_0=\frac{hc}{W_0}\\\text{Where,}\space\lambda_0=\text{threshold wavelength}\\\text{c = velocity of light in vacuum =} 3 × 10^8 \space\text{m/s}\\\text{h = Planck’s constant =} 6.626×10^{-34}\text{Js}\\\text{Substituting the values, we get,}\\\lambda_0=\frac{(6.626×10^{-34}\text{Js})(3 × 10^{8}\text{ms}^{-1})}{1.9×1.602×10^{-19}\text{J}}$$

= 6.53 × 10–7 m

Therefore, the threshold wavelength is 653 nm.

(b) Threshold frequency

From the expression for W0,

$$\text{W}_0=\frac{hc}{\lambda_0}\\\text{or}\space=hv_0\space\bigg(\frac{c}{\lambda}=v_0\bigg)\\v_0=\frac{\text{W}_0}{h}\\=\frac{(1.9×1.602×10^{\normalsize-19}\space\text{J})}{(6.626×10^{\normalsize-34}\text{Js})}$$

Hence, the threshold frequency of radiation is 4.593 × 1014 s–1.

(c) From the equation of Kinetic energy,

$$\text{Kinetic energy = h(v – v}_0) = \text{hc}\bigg(\frac{1}{\lambda}-\frac{1}{\lambda_0}\bigg)\\=(6.626×10^{-34})(3×10^{8})\bigg(\frac{\lambda_0-\lambda}{\lambda_0\lambda}\bigg)\\=(6.626×10^{-34})(3×10^{8})\frac{(653×10^{-9}-500×10^{\normalsize-9})}{(653×10^{\normalsize-9})×(500×10^{\normalsize-9})}\\=\frac{(1.987×10^{\normalsize-26})(153×10^{9})}{653×500}$$

= 9.3149 × 10–20 J

$$\text{We know that K.E}=\frac{1}{2}\text{mv}^{2}\\v=\sqrt{\frac{2\text{K.E.}}{m}}\\=\sqrt{\frac{2(9.3149×10^{\normalsize-20})}{(9.1×10^{\normalsize-31}\text{kg})}}\\v = 4.52 × 10^5 \text{ms}^{\normalsize–1}$$

2.52. Following results are observed when sodium metal is irradiated with different wavelengths. Calculate (a) threshold wavelength and, (b) Planck’s constant.

 λ(nm) = 500 450 400 v × 10–5 (cm s–1) = 2.55 4.35 5.35

$$\textbf{Ans.}\space\text{(a) Threshold wavelength}\\\text{KE}=hc\bigg(\frac{1}{\lambda}-\frac{1}{\lambda_0}\bigg)=\frac{1}{2}mv^{2}\\hc\bigg(\frac{1}{500×10^{\normalsize-9}}-\frac{1}{\lambda_0×10^{\normalsize-9}}\bigg)=\frac{1}{2}\text{m}\\(2.55×10^{+5}×10^{-2})^{2}\\\text{or},\space\frac{hc}{10^{\normalsize-9}}\bigg(\frac{1}{500}-\frac{1}{\lambda_0}\bigg)=\frac{1}{2}\text{m}\space(2.55×10^{+3})^{2}\\\text{…(i)}\\\text{Similarly, for 450 nm}\\\frac{hc}{10^{\normalsize-9}}\bigg(\frac{1}{400}-\frac{1}{\lambda_0}\bigg)=\frac{1}{2}\text{m}(5.35×10^{+3})^{2}\space\space…(iii)\\\text{Dividing equation (iii) by (i),}\\\frac{hc}{10^{\normalsize-9}}\bigg(\frac{1}{400}-\frac{1}{\lambda_0}\bigg)=\frac{1}{2}\text{m}(5.35×10^{+3})^{2}\frac{hc}{10^{\normalsize-9}}\bigg(\frac{1}{500}-\frac{1}{\lambda_0}\bigg)\\=\frac{1}{2}\text{m}(2.55×10^{+3})^{2}$$

$$\frac{\bigg[\frac{\lambda_0-400}{400\lambda_0}\bigg]}{\bigg[\frac{\lambda_0-500}{500\lambda_0}\bigg]}=\frac{(5.35×10^{+3}\text{ms}^{-1})^{2}}{(2.55×10^{+3}\text{ms}^{-1})^{2}}\\\bigg(\frac{5\lambda_0-2000}{4\lambda_0-2000}\bigg)=\bigg(\frac{5.35}{2.55}\bigg)^{2}\\17.60 \lambda_0 – 5\lambda_0 = 8803.537 – 2000\\\lambda_0=\frac{6805.537}{12.607}$$

= 540 (approx.)

Now, for Plancks Constant,

For given values of velocity, h cannot be determined.

2.53. The ejection of the photoelectron from the silver metal in the photoelectric effect experiment can be stopped by applying the voltage of 0.35 V when the radiation 256.7 nm is used. Calculate the work function for silver metal.

Ans. We know that,

E = W0 + K.E

W0 = E – K.E

$$\text{Now, Energy of incident photon (E) is given by}=\frac{hc}{\lambda}$$

Where, c = velocity of light in vacuum

= 3 × 108 m/s

h = Planck’s constant = 6.626 × 10–34 Js

λ = wavelength of radiation = 256.7 × 10-9 m

Substituting the values in the given expression of E:

$$\text{E}=\frac{(6.626×10^{\normalsize-34}\text{Js})(3×10^{8}\text{ms}^{\normalsize-1})}{256.7×10^{-9}\text{m}}\\=7.74×10^{\normalsize-19}\text{J}\\=\frac{7.74×10^{\normalsize-19}}{1.602×10^{\normalsize-19}\text{eV}}$$

or, E = 4.83 eV

Now, the potential applied to silver metal changes to kinetic energy (K.E) of the photoelectron.

Hence, K.E = 0.35 V

K.E = 0.35 eV

Work function, W0 = E – K.E

= 4.83 eV – 0.35 eV

= 4.48 eV

2.54. If the photon of the wavelength 150 pm strikes an atom and one of its inner bound electrons is ejected out with a velocity of 1.5 × 107 ms–1, calculate the energy with which it is bound to the nucleus.

Ans. Energy of incident radiation = Binding energy of electron + Kinetic energy of ejected electron

∴ Binding energy of electron = Energy of incident radiation – Kinetic energy of electron

Energy of incident photon (E) is given by,

$$\text{E}=\frac{hc}{\lambda}$$

Where,

c = velocity of light in vacuum = 3 × 108m/s

h = Planck’s constant = 6.626 × 10–34 Js

λ = wavelength of radiation = 150 pm = 150 × 10–12 m

Substituting the values in the given expression of E:

$$\text{E}=\frac{(6.626×10^{\normalsize-34})(3×10^{8})}{(150×10^{\normalsize-12})}\\= 1.32 × 10^{–15} \text{J}\\\text{Now, energy of ejected electron = KE =}\frac{1}{2}\text{mv}^{2}\\=\frac{1}{2}(9.109×10^{\normalsize-31}\text{kg})(1.5×10^{7})^{2}$$

= 10.24 × 10–17 J

= 1.024 × 10–16 J

Hence, the energy with which the electron is bound to the nucleus can be obtained as:

= E – K.E

= (1.32 × 10–15) J – (1.024× 10–16) J

= 12.22 × 10–16 J

$$\text{or},\space=\frac{12.22×10^{16}}{(1.602×10^{19})}\text{eV}\\= 7.6 × 10^3 \text{eV}$$

2.55. Emission transitions in the Paschen series end at orbit n = 3 and start from orbit n and can be represented as v = 3.29 × 1015 (Hz) $$\bigg[\frac{1}{3^{2}}-\frac{1}{n^{2}}\bigg].$$

Calculate the value of n if the transition is observed at 1285 nm. Find the region of the spectrum.

Ans. Given that,

Wavelength of transition = 1285 nm

= 1285 × 10–9 m

$$v=3.29 × 10^{15} (\text{Hz})\bigg[\frac{1}{3^{2}}-\frac{1}{n^{2}}\bigg]\\\frac{2.3346}{32.9}=\frac{1}{3^{2}}-\frac{1}{n^{2}}\\\text{or}\space 0.071=\frac{1}{9}-\frac{1}{n^{2}}\\\frac{1}{n^{2}}=\frac{1}{9}-0.071\\= 0.111 – 0.071 = 0.04\\n^{2}=\frac{1}{0.0.4}=25\space\text{or n = 5}$$

2.56. Calculate the wavelength for the emission transition if it starts from the orbit having radius 1.3225 nm and ends at 211.6 pm. Name the series to which this transition belongs and the region of the spectrum.

Ans. The radius of the nth orbit of hydrogen-like particles is given by,

$$r = 0.529\frac{n^{2}}{Z}Å$$

= 52.9 n2/Z pm

For radius (r1) = 1.3225 nm

= 1.32225 × 10–9 m

= 1322.25 × 10–12 m

= 1322.25 pm

$$n_1^{2}=\frac{r_1Z}{52.9}\\=\frac{1322.25Z}{52.9}\space\text{...(i)}\\\text{Similarly,}\\n_2^{2}=\frac{r_2Z}{52.9}\\=\frac{211.6Z}{52.9}\space\text{...(ii)}\\\text{Dividing equation (i) and (ii),}\\\frac{n_1^{2}}{n_2^{2}}=\frac{1322.2}{211.6}\\= 6.25\\\frac{n_1}{n_2}=2.5\\\frac{n_1}{n_2}=\frac{25}{10}\text{or}\frac{5}{2}$$

Thus, n1 = 5 and n2 = 2. i.e., the transition is from the 5th orbit to the 2nd orbit. It belongs to the Balmer series. It is in visible region of spectrum. Now, wavenumber is given by:

$$=1.097×10^{7}\bigg(\frac{1}{2^{2}}-\frac{1}{5^{2}}\bigg)\\= 2.303 × 10^6 \text{m}^{–1}\\\text{Wavelength (λ) associated with the emission}\\\text{transition =}\frac{1}{\text{wave number}}\\=\frac{1}{2.303×10^{6}}\\= 0.434 ×10^{–6} \text{m}\\\lambda=434\space\text{nm}$$

2.57. Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 × 106 ms–1, calculate de Broglie wavelength associated with this electron.

Ans. According to de Broglie’s equation,

$$\lambda=\frac{h}{mv}\\=\frac{(6.626×10^{\normalsize-34}\text{Js})}{(9.109×10^{\normalsize-31}\text{kg})(1.6×10^{6}\text{ms}^{\normalsize-1})}$$

= 4.55 × 10–10 m

or, λ = 455 pm

Thus, the de Broglie wavelength associated with the electron is 455 pm.

2.58. Similar to electron diffraction, neutron diffraction microscope is also used for the determination of the structure of molecules. If the wavelength used here is 800 pm, calculate the characteristic velocity associated with the neutron.

Ans. l = 800 pm or 800 × 10–12 m = 8 × 10–10 m

(Given)

According to de Broglie’s equation,

$$\lambda=\frac{h}{mv}\space(\text{m = mass of neutron})\\\text{or},\space v=\frac{h}{m\lambda}\\v=\frac{(6.626×10^{\normalsize-34})}{(1.67×10^{\normalsize-27})(800×10^{\normalsize-12})}$$

= 4.94 × 102 ms–1

or v = 494 ms–1

Therefore, the velocity associated with the neutron is 494 ms–1

2.59. If the velocity of the electron in Bohr’s first orbit is 2.19 × 106 ms–1, calculate the de Broglie wavelength associated with it.

Ans. According to de Broglie’s equation,

$$\lambda=\frac{h}{mv}\space(\text{m = mass of electron})\\=\frac{6.626×10^{\normalsize-34}}{(9.109×10^{\normalsize-31})(2.19×10^{6})}$$

= 332 × 10–12 m

λ = 332 pm

Thus, de Broglie wavelength is 332.43 pm.

2.60. The velocity associated with a proton moving in a potential difference of 1000 V is 4.37 × 105 ms–1. If the hockey ball of mass 0.1 kg is moving with this velocity, calculate the wavelength associated with this velocity.

Ans. According to de Broglie’s equation,

$$\lambda=\frac{h}{mv}\\\text{where h = Planck’s constant}\\= 6.626 × 10^{–34} \text{kg}\space \text{m}^2\text{s}^{\normalsize–1}\\\text{m = mass of hockey ball = 0.1 kg (Given)}\\\text{v = velocity = 4.37 × 105 ms–1 (Given)}\\=\frac{(6.626×10^{\normalsize-34})}{(0.1)(4.37×10^{5})}$$

λ = 1.51 × 10–38 m

Therefore, wavelength = 1.516 × 10–38 m.

2.61. If the position of the electron is measured within an accuracy of + 0.002 nm, calculate the uncertainty in the momentum of the electron. Suppose the momentum of the electron is $$\frac{h}{4\pi_m×0.05\space\text{nm}}$$, is there any problem in defining this value.

Ans. According to Heisenberg’s uncertainty principle,

$$\Delta x×\Delta p=\frac{h}{4\pi}\Rarr\Delta p=\frac{1}{\Delta x}.\frac{h}{4\pi}$$

Where,

∆x = uncertainty in position of the electron
∆p = uncertainty in momentum of the electron
∆x = uncertainty in position of the electron
∆p = uncertainty in momentum of the electron
Substituting the values in the expression of ∆p:

$$\Delta p=\frac{1}{2×10^{\normalsize-12}}×\frac{6.626×10^{\normalsize-34}}{4×3.14}\\= 2.637 × 10^{\normalsize–23} \text{Jsm}^{\normalsize–1}\\\text{Thus, uncertainty in momentum}\\=2.637 × 10^{\normalsize–23} \text{kgms}^{\normalsize–1}\\\text{But actual momentum}=\frac{h}{4\pi_m×0.05\space\text{nm}}(\text{Given})\\=\frac{(6.626×10^{\normalsize-34})}{(4×3.14×5.0×10^{\normalsize-11})}$$

= 1.055 × 10–24 kgms–1

The magnitude of the actual momentum is smaller than the uncertainty, the value cannot be defined.

2.62. The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:

$$\text{1. n = 4, l = 2, m}_l = –2 , m_s =-\frac{1}{2}\\\text{2. n = 3, l = 2, m}_l = 1 , m_s =+\frac{1}{2}\\\text{3. n = 4, l = 1, m}_l = 0, m_s=+\frac{1}{2}\\4.\space n = 3, l = 2, m_l = – 2, m_s=-\frac{1}{2}\\\text{5. n = 3, l = 1, ml = –1, ms = +}\frac{1}{2}\\6.\space n = 4, l = 1, ml = 0, ms =+\frac{1}{2}$$

Ans. 1. For n = 4 and l = 2, occupied orbital is 4d.

2. For n = 3 and l = 2, occupied orbital is 3d.

3. For n = 4 and l = 1 occupied orbital is 4p.

4. For n = 3 and l = 2 occupied orbital is 3d.

5. For n = 3 and l = 1 occupied orbital is 3p.

6. For n = 4 and l = 1, occupied orbital is 4p.

Hence, the six electrons i.e., 1, 2, 3, 4, 5, and 6 are present in the 4d, 3d, 4p, 3d, 3p, and 4p orbitals respectively.

Thus, the increasing order of energies is 5(3p) < 2(3d) = 4(3d) < 3(4p) = 6(4p) < 1 (4d).

2.63. The bromine atom possesses 35 electrons. It contains 6 electrons in 2p orbital, 6 electrons in 3p orbital and 5 electrons in 4p orbital. Which of these electron experiences the lowest effective nuclear charge?

Ans. The factors on which effective nuclear charge experienced by an electron depends is the distance between the nucleus and the orbital, in which the electron is present and the shielding effect of the inner orbital electrons. Effective nuclear charge decreases as distance increases.

Among the given orbitals, 4p orbitals are farthest from the nucleus.Therefore the electrons present in this orbital will experience the lowest effective nuclear charge. Moreover, these electrons are also shielded by electrons present in the 2p and 3p orbitals along with the s-orbitals. Thus, the electrons present in 4p orbitals, will experience the lowest nuclear charge.

2.64. Among the following pairs of orbitals which orbital will experience the larger effective nuclear charge? (i) 2s and 3s, (ii) 4d and 4f,

(iii) 3d and 3p.

Ans. As we know, that greater the penetration of the electron present in a particular orbital towards the nucleus, more will be the magnitude of the effective nuclear charge.

(i) 2s and 3s: The electrons present in the 2s orbital will experience greater nuclear charge than the electron(s) in the 3s orbital because they are closer to nucleus.

(ii) 4d and 4f : 4d will experience greater nuclear charge than 4f as 4d is closer to the nucleus.

(iii) 3d and 3f : 3d will experience greater nuclear charge since it is closer to the nucleus than 3f.

2.65. The unpaired electrons in Al and Si are present in 3p orbital. Which electrons will experience more effective nuclear charge from the nucleus?

Ans. In both the atoms, i.e., Al and Si, the unpaired electrons are present in the 3p orbital with same value of n and 1 but since the atomic number of silicon (Z = 14) is greater than the atomic number of aluminium (Z = 13). Therefore, silicon has a larger nuclear charge than aluminium. Thus, the electrons in the 3p orbital of silicon will experience a more effective nuclear charge than aluminium.

2.66. Indicate the number of unpaired electrons in: (a) P, (b) Si, (c) Cr, (d) Fe and (e) Kr.

Ans. (a) Phosphorous (P)

Atomic number (Z) = 15

The electronic configuration of P is: 1s2 2s2 2p6 3s2 3p3

Thus, phosphorous (P), has three unpaired electrons.

(b) Silicon (Si)

Atomic number (Z) = 14

The electronic configuration of Si is: 1s2 2s2 2p63s2 3p2

Thus, silicon (Si) has two unpaired electrons.

(c) Chromium (Cr)

Atomic number (Z) = 24

The electronic configuration of Cr is: 1s2 2s2 2p6 3s2 3p6 4s13d5

Thus, chromium (Cr) has six unpaired electrons.

(d) Iron (Fe)

Atomic number (Z) = 26

The electronic configuration of Fe is: 1s2 2s2 2p6 3s2 3p6 4s23d6

Thus, iron (Fe) has four unpaired electrons.

(e) Krypton (Kr)

Atomic number (Z) = 36

The electronic configuration of Kr is: 1s2 2s2 2p6 3s2 3p6 4s23d104p6

Since all orbitals are fully occupied, there are no unpaired electrons in krypton (Kr) because all the orbitals are fully occupied.

2.67. (a) How many sub-shells are associated with n = 4?

(b) How many electrons will be present in the sub-shells having ms value of

$$-\frac{1}{2}$$

for  n = 4

Ans. (a) Given,

n = 4

For a given value of ‘n’, ‘l’ can have values from zero to (n – 1).

l = 0, 1, 2, 3

Thus, four sub-shells are associated with n = 4, which are s, p, d and f.

(b) Number of orbitals in the nth shell = n2

For n = 4

Total number of orbitals will be 16

If each orbital is taken fully, then it will have 1 electron with ms value of
$$-\frac{1}{2}$$

Therefore, total number of electrons with ms value of

$$-\frac{1}{2}$$

is 16.