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 in $3n$, the maximum velocity of the ejected photoelectrons will be :

• A. Equal to $\sqrt{3}v$
• B. Equal to $2v$
• C. More than $\sqrt{3}v$
• D. Less than $\sqrt{3}v$

Answer 1

Answer: (C)

More than $\sqrt{3}v$

$E=\frac{1}{2}m{V}_0^2+{\varphi }_o$

$hn=\frac{1}{2}m{V}^2+{\varphi }_o.......\left(1\right)$

$h\times 3n=\frac{1}{2}m{v}_1^2+{\varphi }_o.........\left(2\right)$

From equations 1 and 2.

$3\times \frac{1}{2}m{v}^2+3{\varphi }_o=\frac{1}{2}m{v}_1^2+{\varphi }_o$

$v_1^2=3v^2+\frac{4{\varphi }_o}{m}$

$v_1^2>3v^2$

$v_1>\sqrt{3}v$