BITSAT 2015 Paper

As we know, threshold wavelength
(λ0) =

hc

Φ

⇒ λ0 =

(6.63×10−34)×3×108

2.3×(1.6×10−19)

= 5.404 × 10−7 m
⇒ λ0 = 5404

0

A

Hence, wavelength 4144 Å and 4972 Å will emit electron from the metal surface.
For each wavelength energy incident on the surface per unit time = intensity of each × area of the surface wavelength
=

3.6×10−3

3

× (1‌cm)2 = 1.2 × 10−7 joule
Therefore, energy incident on the surface for each wavelength in 2s
E = (1.2 × 10−7) × 2 = 2.4 × 10−7 J
Number of photons n1 due to wavelength 4144 Å
n1 =

(2.4×10−7)(4144×10−10)

(6.63×10−34)(3×108)

= 0.5 8 1012
Number of photons n2 due to the wavelength 4972 Å
n2 =

(2.4×10−7)(4972×10−10)

(6.63×10−34)(3×108)

= 0.572 × 1012
Therefore total number of photo electrons liberated in 2s,
N = n1+n2
= 0.5 × 1012 + 0.575 × 1012
= 1.075 × 1012