Question

Define the term "cut off frequency" in photoelectric emission. The threshod frequency of a metal is f. When the light of frequency 2f is incident on the metal plate, the maximum velocity of photo-electrons is v₁. When the frequency of the incident radiation is increased to 5f, the maximum velocity of phto-electrons is v₂. Find the ratio v₁ : v₂.

Answer 1

That particular frequency of incident radiation for which the minimum negative potential V₀ given to a plate for photelectric current becomes zero is called the cut-off frequency. 
Given that threshold frequency of metal is f. For light of frequency 2f, using Einstein's equation for photoelectric effect, we can write

\[h\left( 2f - f \right) = \frac{1}{2}m {v_1}^2\] .......... (1)

Similarly, for light having frequency 5f,  we have

\[h\left( 5f - f \right) = \frac{1}{2}m {v_2}^2\] .........(2)

Using (1) and (2), we find

\[\frac{f}{4f} = \frac{{v_1}^2}{{v_2}^2} \Rightarrow \frac{v_1}{v_2} = \sqrt{\frac{1}{4}}\]

\[ \Rightarrow \frac{v_1}{v_2} = \frac{1}{2}\]