Question

For a certain metal $v$ is five times the $v_0$ and the maximum velocity of coming photo-electrons is $8\times {10}^6$. If $v=2v_0$, the maximum velocity of photo electrons will be :

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

Answer 1

The correct option is A $4\times {10}^{6}m/s$

Given the frequency of excitation is given as : ${\nu }_0$

Given initial frequency $=\nu =5{\nu }_0$

$5h{\nu }_0-h{\nu }_0=\frac{1}{2}m{v}_1^2=4h{\nu }_0$

Given that final frequency $=\nu =2{\nu }_0$

$2h{\nu }_0-h{\nu }_0=\frac{1}{2}m{v}_2^2=h{\nu }_0$

Dividing the two:

$\frac{v_2^2}{v_1^2}=\frac{1}{4}$

$v_2=\frac{1}{2}{v}_1=\frac{1}{2}8\times {10}^6=4\times {10}^{6}m/s$

Option A is correct.