Question

Two photons of energy $2.5eV$ and $3.5eV$ fall on a metal surface of work function $1.5eV$. The ratio of the maximum velocities of the photoelectrons emitted from the metal surface is:

• A. $1:4$
• B. $2:1$
• C. $1:2$
• D. $1:\sqrt{2}$

Answer 1

The correct option is D $1:\sqrt{2}$

Energy of Ist photon $E_1=2.5eV$

Energy of second photon $E_2=3.5eV$

Work function of metal $\varphi =1.5eV$

Let the maximum velocities of the photoelectrons emitted be $v_1$ and $v_2$ respectively.

Using $K.E=E-\varphi$ where $K.E=\frac{1}{2}{m}_e{v}^2$

$\therefore$ $\frac{1}{2}{m}_e{v}^2=E-\varphi$ ........(1)

$⟹$ $\frac{v_1^2}{v_2^2}=\frac{E_1-\varphi }{E_2-\varphi }$

OR $\frac{v_1^2}{v_2^2}=\frac{2.5-1.5}{3.5-1.5}=\frac{1}{2}$ $⟹v_1:v_2=1:\sqrt{2}$