Question

Light of two different frequency whose photons have energies of 1 eV and 2.5 eV are incident one by one on a metal surface of work function 0.5 eV. The ratio of maximum energies of emitted electrons will be

• A. $1:4$
• B. $4:1$
• C. $1:2$
• D. $2:1$

Answer 1

The correct option is C $1:2$

For maximum speed of the photo electrons $\frac{1}{2}m{v}^2=E_{photon}-\varphi$

where $\varphi =0.5eV$ is the work function of the metal.

Energy of photon $E_1=1eV$

$\therefore \frac{1}{2}mv{}_1^2=1-0.5$

We get $\frac{1}{2}mv{}_1^2=1-\mathrm{0.5.......}\left(1\right)$

Energy of photon $E_2=2.5eV$

$\therefore \frac{1}{2}mv{}_2^2=2.5-0.5$

We get $\frac{1}{2}mv{}_2^2=2eV....\left(2\right)$

Ratio of maximum speeds $\frac{v{}_1^2}{v{}_2^2}=\frac{0.5}{2}=\frac{1}{4}$

we get $\frac{v_1}{v_2}=\frac{1}{2}$