Question

The maximum velocity of electrons emitted from a metal surface is $V$. When frequency of light falling on it is $f$. The maximum velocity when frequency becomes $4f$ is

• A. $>2V$
• B. $<2V$
• C. Between $2V$ 𝑎𝑛𝑑 $4V$
• D. $2V$

Answer 1

The correct option is A $>2V$

According to Einstein’s photoelectric equation

Maximum kinetic energy

$K_{max}=hf-{\varphi }_o$

$\Rightarrow \frac{1}{2}m{v}^2=hf-{\varphi }_0$

For frequency $f$, maximum velocity $v$

$\Rightarrow \frac{1}{2}m{v}^2=hf-{\varphi }_o\dots \left(1\right)$

For frequency $4f$, let maximum velocity $=v_1$

$\Rightarrow \frac{1}{2}m{v}_1^2=4hf-{\varphi }_o\dots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$\Rightarrow \left(\frac{v_1^2}{v^2}\right)=\frac{4hf-{\varphi }_o}{hf-{\varphi }_o}$

$\Rightarrow {\left(\frac{v_1}{v}\right)}^2=\frac{4-\left(\frac{{\varphi }_0}{hf}\right)}{1-\left(\frac{{\varphi }_o}{hf}\right)}$

Since ${\varphi }_o\ne 0$

$\Rightarrow {\left(\frac{v_1}{v}\right)}^2>4$

$\Rightarrow v_1>2v$

Hence, option (𝒃) is correct.