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)

$E=\frac{1}{2}m{v}_0^2+{\varphi }_o$
$hn=\frac{1}{2}m{v}^2+{\left(\varphi \right)}_0...\left(1\right)$
$h\times 3n=\frac{1}{2}m{v}_1^2+{\left(\varphi \right)}_0...\left(2\right)$
From equations $\left(1\right)$ and $\left(2\right)$ ,
$3\times \frac{1}{2}m{v}^2+3{\varphi }_0=\frac{1}{2}m{v}_1^2+{\varphi }_0$
$v_1^2=3v^2+\frac{4{\varphi }_o}{m}$
$v_1^2>3v^2$
$v_1>\sqrt{3}v$ .