**Question 2.9: A photon of wavelength 4 × 10 ^{–7} m strikes on the metal surface, the work function of the metal is 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).

**Question 2.9: A photon of wavelength 4 × 10 ^{–7} m strikes on the metal surface, the work function of the metal is 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).**Ans 2.9:**(i) We know

Energy (E) of a photon = hν

E = hc/Λ ————-(1) (ν = c/Λ)

Where, h = Planck’s constant = 6.626 × 10

^{–34}Js

c = velocity of light in vacuum = 3 × 10

^{8}m/s

λ = wavelength of photon = 4 × 10

^{–7}m

Substituting the values in relation (1)

E = 6.626 × 10

^{–34}Js x 3 × 10

^{8}m/s/ 4 × 10

^{–7}m

Hence, the energy of the photon is 4.97 × 10

^{–19}J.

The energy of the photon (eV),

E = 4.97 × 10

^{–19}/1.6020 × 10

^{–19}J = 3.10 ev

(ii) The kinetic energy of emission E_{k} is given by

KE= hν – hν_{0} ————(2)

But we know E = hν, w_{0} = hν_{0}

KE = E – w_{0}

KE = 3.10 – 2.13 = 0.97 ev

(iii) The velocity of a photoelectron (ν)

KE = 1/2mv^{2}

KE = 0.97 ev x 1.6020 × 10^{–19} J

0.97 ev x 1.6020 × 10^{–19} J = ½ 9.1 x 10^{-31} kg x v^{2}

We know

1J = Kgm^{2}s^{-2}

On solving

v^{2 }= 34.12 x 10^{10} m^{2}s^{-2}

v = 5.8 x 10^{5} m/s