Question

The surface of a metal is illuminated with the light of $ 400nm $. The kinetic energy of the ejected photoelectrons was found to be $ 1.68{\text{eV}} $. The work function of the metal is:
 $ \left( {hc = 1240{\text{eV}}{\text{.nm}}} \right) $
(A) $ 3.09{\text{eV}} $
(B) $ 1.41{\text{eV}} $
(C) $ 1.51{\text{eV}} $
(D) $ 1.68{\text{eV}} $

Answer 1

Hint
To solve this question, we need to use the Einstein photoelectric equation. The two values out of the three quantities involved are already given in the question. So, the third unknown quantity, the work function can be easily found out.
Formula Used: The formulae used in solving this question are given by:
 $\Rightarrow {K_{\max }} = h\nu - {\varphi _0} $
Here $ {K_{\max }} $ is the maximum kinetic energy of the ejected photoelectrons, when a light of frequency $ \nu $ is incident upon a metal having a work function of $ {\varphi _0} $.
 $\Rightarrow \nu = \dfrac{c}{\lambda } $
Here $ \nu $ is the frequency of a light wave having a speed of $ c $ and wavelength of $ \lambda $.

Complete step by step answer
As we know that the Einstein’s photo-electric equation is given by
 $\Rightarrow {K_{\max }} = h\nu - {\varphi _0} $ (1)
Also we know that the frequency of a light wave is given by
 $\Rightarrow \nu = \dfrac{c}{\lambda } $ (2)
Substituting (2) in (1), we get
 $\Rightarrow {K_{\max }} = \dfrac{{hc}}{\lambda } - {\varphi _0} $
So the work function of the metal is given by
 $\Rightarrow {\varphi _0} = \dfrac{{hc}}{\lambda } - {K_{\max }} $ (3)
According to the question, $ \lambda = 400nm $, $ {K_{\max }} = 1.68{\text{eV}} $, and $ hc = 1240{\text{eV}}{\text{.nm}} $
Putting these values in (3) we get
 $\Rightarrow {\varphi _0} = \dfrac{{1240}}{{400}} - 1.68 $
 $\Rightarrow {\varphi _0} = 1.42{\text{eV}} $
Thus the work function of the metal comes out to be equal to $ 1.42{\text{eV}} $.
Looking at the options, we find the nearest value to be $ 1.41{\text{eV}} $.
Hence, the correct answer is option B.

Note
The kinetic energy term in the Einstein photoelectric equation always corresponds to the maximum kinetic energy of the ejected photoelectrons. The electrons which are present at the outermost layer of the metal possess this maximum value. In this question, it is not mentioned that the value of the kinetic energy given is the maximum. But we would ourselves take it to be the maximum, to find out the work function of the metal.