Question

Threshold frequency for a certain metal is $v_0$. When light of frequency $2v_0$ 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 $5v_0$, 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
$hv-h{v}_0=\frac{1}{2}m{v}_{\max }^2$
When $v=2v_0,v_{\max }=4\times 10^{8}cm{s}^{-1}$
$2h{v}_0-h{v}_0=(1/2)m{\left(4\times 10^8\right)}^2$
$h{v}_0=\frac{1}{2}m{\left(4\times 10^8\right)}^2$ ...(i)
When $v=5v_0,v_{\max }=v^{\prime }$
$h\left(5v_o\right)-h{v}_o=\frac{1}{2}m{v}^{{}^{\prime }2}$ ...(ii)
Dividing Eq. (ii) by Eq. (i), we get
$v^{\prime }=8\times 10^{8}cm{s}^{-1}$