Question

Light of two different frequencies whose photons have energies $1eV$ and $2.5eV$ respectively, successively illuminate a metallic surface whose work function is $0.5eV$. Ratio of maximum speeds of emitted electrons will be :

• A. $1:4$
• B. $1:1$
• C. $1:5$
• D. $1:2$

Answer 1

The correct option is D $1:2$

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is

$K_{max}=hv-{\varphi }_0$

where $hv$ is the energy of incident photon and ${\varphi }_0$ is the work function.

But $K_{max}=\frac{1}{2}m{v}_{max}^2$

$\therefore \frac{1}{2}m{v}_{max}^2=hv-{\varphi }_0$

As per question,

$\frac{1}{2}m{v}_{ma{x}_1}^2=1eV-0.5eV=0.5eV....\left(i\right)$

and $\frac{1}{2}m{v}^{2}ma{x}_2=2.5eV=0.5eV=2eV.....\left(ii\right)$

Dividing eqn. (i) by eqn. (ii), we get

$\frac{v_{ma{x}_1}^2}{v_{ma{x}_2}^2}=\frac{0.5eV}{2eV}=\frac{1}{4}$

$\frac{v_{ma{x}_1}}{v_{ma{x}_2}}=\sqrt{\frac{1}{4}}=\frac{1}{2}$.