Question

The threshold frequency for certain metal is $v_0$. The light of frequency $2v_0$ is incident on it, the maximum velocity of photoelectrons is $4\times {10}^{6}m{s}^{-1}$ . If the frequency of incident radiation is increased to $5v_0$ , the maximum velocity of photoelectrons will be (in m/s):

• A. $2\times {10}^7$
• B. $2\times {10}^6$
• C. $8\times {10}^6$
• D. $\frac{4}{5}\times {10}^6$

Answer 1

Answer: (C)

When light of frequency $2v_o$ is incident,

$\frac{1}{2}m(4\times {10}^6{)}^2=h(2v_o-v_o)$

$\frac{1}{2}m(4\times {10}^6{)}^2=h{v}_o$

When light of frequency $5v_o$ is incident,

$\frac{1}{2}m{u}^2=h(5v_o-v_o)=4h{v}_o$

Solving above equations, $u=8\times {10}^{6}m/s$