Question

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

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

Answer 1

The correct option is D

$8\times {10}^{6}m{s}^{-1}$

Step 1: Given parameters

Threshold frequency for metal = $v_0$

The initial frequency of incident light $v_1=2v_0$

Initial maximum velocity of photoelectrons $V_1=4\times {10}^{6}m{s}^{-1}$

Final frequency of incident light $v_2=5v_0$

Step 2: To find

Final maximum velocity of photoelectrons $V_2=?$

Step 3: Calculation

Einstein’s equation for the photoelectric effect is given by,

$KE=hv-h{v}_0$

where,

$h$= Planck’s constant

$v_0$= threshold frequency

$V$= frequency of incident light

$KE$= maximum kinetic energy of an electron

and $KE=\frac{1}{2}m{V}^2$

Case 1:

$KE=\frac{1}{2}m{V_1}^2$

$\frac{1}{2}m{V_1}^2=h{v}_1-h{v}_0$

$\frac{1}{2}m{V_1}^2=2h{v}_0-h{v}_0$

$\frac{1}{2}m{V_1}^2=h{v}_0$ -------------------------------[1]

Case 2:

$KE=\frac{1}{2}m{V_2}^2$

$\frac{1}{2}m{V_2}^2=h{v}_2-h{v}_0$

$\frac{1}{2}m{V_2}^2=5h{v}_0-h{v}_0$

$\frac{1}{2}m{V_2}^2=4h{v}_0$ ------------------------------[2]

dividing equation [2] by equation [1], we get

$V_2=2V_1$

$V_2=2\times 4\times {10}^{6}m{s}^{-1}$

$V_2=8\times {10}^{6}m{s}^{-1}$

The maximum velocity of photoelectrons will be $8\times {10}^{6}m{s}^{-1}$.

Therefore, the correct option is D.