Q116 / 155 · WBJEE 2026 · Physics
Radiation of wavelength $\lambda$ is incident on a photocell. The fastest emitted electron has speed $v$. If the wavelength is changed to $\dfrac{3\lambda}{4}$, then the speed of the fastest emitted electron will be –
- Correct answerAgreater than $v\sqrt{\dfrac{4}{3}}$
- Bless than $v\sqrt{\dfrac{4}{3}}$
- Cequal to $v\sqrt{\dfrac{4}{3}}$
- Dequal to $v\sqrt{\dfrac{3}{4}}$
Solution
Using Einstein's photoelectric equation:
$\dfrac{1}{2}mv^2 = \dfrac{hc}{\lambda} - \phi$
When the wavelength is changed to $\dfrac{3\lambda}{4}$, the new maximum kinetic energy is:
$\dfrac{1}{2}m(v')^2 = \dfrac{hc}{\dfrac{3\lambda}{4}} - \phi = \dfrac{4hc}{3\lambda} - \phi$
Substituting $\dfrac{hc}{\lambda} = \dfrac{1}{2}mv^2 + \phi$ into the second equation:
$\dfrac{1}{2}m(v')^2 = \dfrac{4}{3}\left(\dfrac{1}{2}mv^2 + \phi\right) - \phi$
$\dfrac{1}{2}m(v')^2 = \dfrac{4}{3}\left(\dfrac{1}{2}mv^2\right) + \dfrac{1}{3}\phi$
Since the work function $\phi > 0$, we get:
$\dfrac{1}{2}m(v')^2 > \dfrac{4}{3}\left(\dfrac{1}{2}mv^2\right)$
$(v')^2 > \dfrac{4}{3}v^2$
$v' > v\sqrt{\dfrac{4}{3}}$
Answer: greater than $v\sqrt{\dfrac{4}{3}}$
$\dfrac{1}{2}mv^2 = \dfrac{hc}{\lambda} - \phi$
When the wavelength is changed to $\dfrac{3\lambda}{4}$, the new maximum kinetic energy is:
$\dfrac{1}{2}m(v')^2 = \dfrac{hc}{\dfrac{3\lambda}{4}} - \phi = \dfrac{4hc}{3\lambda} - \phi$
Substituting $\dfrac{hc}{\lambda} = \dfrac{1}{2}mv^2 + \phi$ into the second equation:
$\dfrac{1}{2}m(v')^2 = \dfrac{4}{3}\left(\dfrac{1}{2}mv^2 + \phi\right) - \phi$
$\dfrac{1}{2}m(v')^2 = \dfrac{4}{3}\left(\dfrac{1}{2}mv^2\right) + \dfrac{1}{3}\phi$
Since the work function $\phi > 0$, we get:
$\dfrac{1}{2}m(v')^2 > \dfrac{4}{3}\left(\dfrac{1}{2}mv^2\right)$
$(v')^2 > \dfrac{4}{3}v^2$
$v' > v\sqrt{\dfrac{4}{3}}$
Answer: greater than $v\sqrt{\dfrac{4}{3}}$
Hence, the correct answer is option: 'A'.