Question

A photon of wavelength $4\times {10}^{-7}m$ strikes on metal surface. The work function of the metal is 2.13 eV. The velocity of the photo electron is:

• A. $5.67\times {10}^6{ms}^{-1}$
• B. $5.84\times {10}^5{ms}^{-1}$
• C. $5.67\times {10}^{-5}{ms}^{-1}$
• D. $5.67\times {10}^{-6}{ms}^{-1}$

Answer 1

The correct option is B $5.84\times {10}^5{ms}^{-1}$

(1) We know $\lambda =4\times {10}^{–7}$ m (given)

$c=3\times {10}^8$

From the equation $E=hv$ or $hc/\lambda$

Where,

h = Planck’s constant $=6.626\times {10}^{–34}$ Js

c = velocity of light in vacuum $=3\times {10}^8$ m/s

$\lambda =$ wavelength of photon $=4\times 10–7$ m

Substituting the values in the given expression of E:

$E=(6.626\times {10}^{-34})(3\times {10}^8)/(4\times {10}^{-7})=4.9695\times {10}^{-19}J$

Hence, the energy of the photon is $4.97\times {10}^{–19}$ J.

(ii) The kinetic energy of emission KE is given by

$KE=hv-h{v}_0$

$KE=(E-W)eV$

$KE=\frac{4.9695\times {10}^{-19}}{1.6020\times {10}^{-19}}eV-2.13eV$

$KE=(3.1020–2.13)eV$

$KE=0.9720eV$

Hence, the kinetic energy of emission is 0.97 eV.

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

$1/2m{v}^2=hv-h{v}_0$

Where, $(hv-h{v}_0)$ is the kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron. Substituting the values in the given expression of v:

$v=(0.3418\times {10}^{12}{m}^2{s}^{-2}{)}^{1/2}$

$v=5.84\times {10}^{5}m{s}^{–1}$

Hence, the velocity of the photoelectron is $5.84\times {10}^{5}m{s}^{–1}$