Question

The threshold frequency for certain metal is $v_0$. When 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 $5v_0$, then the maximum velocity of photoelectrons will be :

• A. $4/5\times {10}^{6}m{s}^{-1}$
• B. $2\times {10}^{6}m{s}^{-1}$
• C. $8\times {10}^{6}m{s}^{-1}$
• D. $2\times {10}^{7}m{s}^{-1}$

Answer 1

The correct option is C $8\times {10}^{6}m{s}^{-1}$

The Einstein's equation for photoelectric effect is given by , $KE=h\nu -h{\nu }_0$ where $KE=$maximum kinetic energy of electron, $h=$ Planck's constant , ${\nu }_0=$ threshold frequency and $\nu =$ frequency of incident light .

Also $KE=\frac{1}{2}m{v}^2$ where $v=$ maximum velocity of electron.

In first case, $\frac{1}{2}m{v}_1^2=h\left(2{\nu }_0\right)-h{\nu }_0=h{\nu }_0....\left(1\right)$ and

In second case, $\frac{1}{2}m{v}_2^2=h\left(5{\nu }_0\right)-h{\nu }_0=4h{\nu }_0....\left(2\right)$

From (1) and (2), $\frac{1}{2}m{v}_2^2=4\times \frac{1}{2}m{v}_1^2$

or $v_2=2v_1=2\times (4\times {10}^6)=8\times {10}^{6}m/s$