Question

Two identical photocathodes receive the light of frequencies $f_1$ and $f_2$ respectively. If the velocities of the photo-electrons coming out are $v_1$ and $v_2$ respectively, then:

• A. $v_1+v_2={\left[\frac{2h}{m}\left(f_1+f_2\right)\right]}^{\frac{1}{2}}$
• B. $v_1-v_2={\left[\frac{2h}{m}\left(f_1-f_2\right)\right]}^{\frac{1}{2}}$
• C. ${v_1}^2+{v_2}^2=\frac{2h}{m}\left(f_1+f_2\right)$
• D. ${v_1}^2-{v_2}^2=\frac{2h}{m}\left(f_1-f_2\right)$

Answer 1

The correct option is D

${v_1}^2-{v_2}^2=\frac{2h}{m}\left(f_1-f_2\right)$

Step 1: Given

Frequency of first photocathode: $f_1$

Frequency of second photocathode: $f_2$

Velocity of photoelectrons from first photocathode: $v_1$

Velocity of photoelectrons from second photocathode: $v_2$

Step 2: Formula Used

$KE=hf-\varphi$

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

Step 3: Find the difference between kinetic energies of the photoelectrons in terms of frequencies

Assume $K_1$ and $K_2$ to be the kinetic energies of the photoelectrons from first and second photocathode respectively and $\varphi$ to be the work function which will be same for both. Calculate the difference between kinetic energies using the formula. $h$ is the Planck's constant.

$\begin{array}{rcl}{K}_1-K_2& =& \left(h{f}_1-\varphi \right)-\left(h{f}_2-\varphi \right)\\ & =& h{f}_1-\varphi -h{f}_2+\varphi \\ & =& h{f}_1-h{f}_2\\ & =& h\left(f_1-f_2\right)\end{array}$

Step 4: Find the difference between kinetic energies of the photoelectrons in terms of velocities

Calculate the difference in kinetic energies using the formula, mass remains same.

$\begin{array}{rcl}{K}_1-K_2& =& \frac{1}{2}m{v_1}^2-\frac{1}{2}m{v_2}^2\\ & =& \frac{1}{2}m\left({v_1}^2-{v_2}^2\right)\end{array}$

Step 5: Find the relation between frequencies and velocities.

Equate both the expressions obtained.

$$\begin{gathered} \frac{1}{2}m\left({v_1}^2-{v_2}^2\right)=h\left(f_1-f_2\right) \\ \Rightarrow v_1^2-\begin{array}{l}{v}_2^2\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{rcl}& & \frac{2h}{m}\end{array}\begin{array}{rcl}& & \left(f_1-f_2\right)\end{array} \end{gathered}$$

Hence, Option (D) is correct. ${v_1}^2\begin{array}{rcl} & & -\end{array}{\begin{array}{rcl} & & v_2\end{array}}^2=\begin{array}{rcl} & & \frac{2h}{m}\end{array}\begin{array}{l}\left(f_1-f_2\right)\end{array}$.