Question

Threshold frequency for a certain metal is $v_{0}$. When light of frequency $2\, v_{0}$ is incident on it, the maximum velocity of photoelectrons is $4 \times 10^{8} cm s ^{-1}$. If frequency of incident radiation is increased to $5\, v_{0}$, then the maximum velocity of photoelectrons, in $cm s ^{-1}$, will be

• A. $(4 / 5) \times 10^{8}$
• B. $2 \times 10^{8}$
• C. $8 \times 10^{8}$
• D. $20 \times 10^{8}$

Answer 1

Answer: (C)

Einstein's equation for photoelectric effect is
$h v-h v_{0}=\frac{1}{2} m v_{\max }^{2}$
When $v=2 v_{0}, v_{\max }=4 \times 10^{8} cm s ^{-1}$
$2 h v_{0}-h v_{0}=(1 / 2) m\left(4 \times 10^{8}\right)^{2}$
$h v_{0}=\frac{1}{2} m\left(4 \times 10^{8}\right)^{2}$ ...(i)
When $v=5 v_{0}, v_{\max }=v'$
$h\left(5 v_{ o }\right)-h v_{ o }=\frac{1}{2} m v^{'2}$ ...(ii)
Dividing Eq. (ii) by Eq. (i), we get
$v'=8 \times 10^{8} cm s ^{-1}$