Question

The maximum velocity of the photoelectrons emitted from the surface is $v$ when light of frequency $n$ falls on a metal surface. If the incident frequency is increased to $3n$ , the maximum velocity of the ejected photoelectrons will be,

• A. less than $\sqrt{3}v$ .
• B. $v$
• C. more than $\sqrt{3}v$ .
• D. equal to $\sqrt{3}v$ .

Answer 1

Answer: (C)

$E=\frac{1}{2}mv_{0}^{2}+\phi_{o}$
$hn=\frac{1}{2}mv^{2}+\left(\phi\right)_{0}...\left(1\right)$
$h\times 3n=\frac{1}{2}mv_{1}^{2}+\left(\phi\right)_{0}...\left(2\right)$
From equations $\left(1\right)$ and $\left(2\right)$ ,
$3\times \frac{1}{2}mv^{2}+3\phi_{0}=\frac{1}{2}mv_{1}^{2}+\phi_{0}$
$v_{1}^{2}=3v^{2}+\frac{4 \phi_{o}}{m}$
$v_{1}^{2}>3v^{2}$
$v_{1}>\sqrt{3}v$ .