Question: Q. 14. Define the term “cut off frequency” in photoelectric emission. The threshold frequency of a metal is $f$. When the light of frequency $2 f$ is incident on the metal plate, the maximum velocity of photoelectrons is $v_{1}$. When the frequency of the incident radiation is increased to $5 f$, the maximum velocity of photo-electrons is $v_{2}$. Find the ratio $v_{1}: v_{2}$.
$\mathrm{R}$ [Foreign I, II, III 2016]
Show Answer
Solution:
Ans. Cut off frequency : It is that minimum frequency of incident radiation below which no photo emission takes place from a photo electric material.
(Alternatively, That minimum frequency of incident radiation at which photons are just emitted with zero kinetic energy.)
$$ \begin{align*} K_{\max } & =h f-W_{0} \ \frac{1}{2} m v_{1}^{2} & =2 h f-h f=h f \end{align*} $$
$$ \begin{array}{rlrl} & & \frac{1}{2} m v_{2}^{2} & =5 h f-h f=4 h f \ \therefore & \frac{v_{1}^{2}}{v_{2}^{2}} & =\frac{1}{4} \ \Rightarrow & \frac{v_{1}}{v_{2}} & =\frac{1}{2} \ & & \text { [CBSE Marking Scheme 2016] } \end{array} $$
