### Question 2.9

NCERT Class XI Chemistry

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).

### Video Explanation

(detailed solution given after this video)

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### Solution in Detail

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

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 $\displaystyle m_e = 9.1 \times 10^{-31}\text{ kg}$

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

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

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