(solved)Question 2.9 of NCERT Class XI Chemistry Chapter 2

$\displaystyle \underline{\underline{\text{(i) Energy of Photon}}}$

Energy E of 1 photon by Planck's equation

$\displaystyle E = \frac{hc}{\lambda} \text{ (J)}$

h, c, $\displaystyle \lambda$, E mean as usual

but (1): $\displaystyle \lambda = 4 \times 10^{-7} \text{ m (given)}$

use (2): $\displaystyle h = 6.6 \times 10^{-34} \text{ Js}$

$\displaystyle \therefore E = \frac{hc}{\lambda}$

$\displaystyle = \frac{(6.6 \times 10^{-34}) \times (3 \times 10^8)}{4 \times 10^{-7}} \text{ J}$

$\displaystyle = 4.95 \times 10^{-19} \text{ J}$

also, $\displaystyle 1 \text{ eV} = 1.6020 \times 10^{-19} \text{ J (given)}$

$\displaystyle \therefore E = \frac{4.95 \times 10^{-19}}{1.6020 \times 10^{-19}} \text{ (eV)}$

$\displaystyle = 3.1 \text{eV }\underline{Ans}$

$\displaystyle \underline{\underline{\text{(ii) KE of Emission}}}$

Let E = energy of photon, $\displaystyle W_o$ = work function, and KE = kinetic energy of emission

by photoelectric effect: $\displaystyle E = W_o + \text{KE} $

$\displaystyle \implies \text{KE} = \text{E} - W_o$

but $\displaystyle W_o = 2.13\text{ eV (given)}$

$\displaystyle E = 3.1\text{ eV (calculated above)}$

$\displaystyle \therefore \text{KE} = 3.1 \text{ eV} - 2.13\text{ eV}$

= 0.97 eV Ans!

$\displaystyle \underline{\underline{\text{(iii) Velocity of photoelectron}}}$

$\displaystyle \text{KE} = \frac12 m_ev^2$

$\displaystyle \implies v = \sqrt{\frac{2\text{KE}}{m_e}}$

but mass of electron $\displaystyle m_e = 9.1 \times 10^{-31}\text{ kg}$

$\displaystyle v = \sqrt{\frac{2 \times 0.97 \times 1.6020 \times 10^{-19} \text{ J}}{9.1 \times 10^{-31}}}$

$\displaystyle =5.84 \times 10^5 \text{ m/s }\underline{ Ans}$