Question

The threshold frequency for a certain photosensitive metal is ${\nu }_0$. When it is illuminated by light of frequency $\nu =2{\nu }_0$, the maximum velocity of photoelectrons is $v_0$. What will be the maximum velocity of the photoelectrons when the same metal is illuminated b a light of frequency $\nu =5{\nu }_0$?

• A. $\sqrt{2}{v}_0$
• B. $2v_0$
• C. $2\sqrt{2}{v}_0$
• D. $4v_0$

Answer 1

The correct option is B $2v_0$

We know that, $\frac{1}{2}m{v}_{max}^2=h(\nu -{\nu }_0)$
For light of frequency $\nu =2{\nu }_0$, we have $\frac{1}{2}m{v}_0^2=h(2{\nu }_0-{\nu }_0)=h{\nu }_0$ .....(i)
For light of frequency $\nu =5{\nu }_0$, we have $\frac{1}{2}m{v}^2=h(5{\nu }_0-{\nu }_0)=4h{\nu }_0$ .....(ii)
Dividing (ii) by (i), $v^2=4v_0^2$
$\Rightarrow v=2v_0$
Hence, the correct choice is (b).