Q132 / 150 · MHT CET 2026 (16 April Shift 2) · Physics
When a light of wavelength '$\lambda$' falls on the emitter of a photosensitive surface, maximum speed of emitted photoelectrons is 'V'. If the incident wavelength is changed to '$2\lambda/3$', maximum speed of emitted photoelectrons will be
- Aless than $V(1.5)^{\frac{1}{2}}$
- Correct answerBgreater than $V(1.5)^{\frac{1}{2}}$
- Cless than $V$
- Dless than $\dfrac{V}{2}$
Solution
According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons is given by
$\dfrac{1}{2}mV^2 = \dfrac{hc}{\lambda} - \phi$
When the incident wavelength is changed to $\dfrac{2\lambda}{3}$, the new maximum kinetic energy is
$\dfrac{1}{2}mV'^2 = \dfrac{hc}{\dfrac{2\lambda}{3}} - \phi = \dfrac{3hc}{2\lambda} - \phi$
Substituting $\dfrac{hc}{\lambda} = \dfrac{1}{2}mV^2 + \phi$ into the above equation, we get
$\dfrac{1}{2}mV'^2 = \dfrac{3}{2}\left(\dfrac{1}{2}mV^2 + \phi\right) - \phi$
$\dfrac{1}{2}mV'^2 = \dfrac{3}{2}\left(\dfrac{1}{2}mV^2\right) + \dfrac{\phi}{2}$
Since the work function $\phi > 0$, it follows that
$\dfrac{1}{2}mV'^2 > \dfrac{3}{2}\left(\dfrac{1}{2}mV^2\right)$
$V'^2 > 1.5 V^2$
$V' > V(1.5)^{\dfrac{1}{2}}$
Thus, the maximum speed of emitted photoelectrons will be greater than $V(1.5)^{\dfrac{1}{2}}$.
$\dfrac{1}{2}mV^2 = \dfrac{hc}{\lambda} - \phi$
When the incident wavelength is changed to $\dfrac{2\lambda}{3}$, the new maximum kinetic energy is
$\dfrac{1}{2}mV'^2 = \dfrac{hc}{\dfrac{2\lambda}{3}} - \phi = \dfrac{3hc}{2\lambda} - \phi$
Substituting $\dfrac{hc}{\lambda} = \dfrac{1}{2}mV^2 + \phi$ into the above equation, we get
$\dfrac{1}{2}mV'^2 = \dfrac{3}{2}\left(\dfrac{1}{2}mV^2 + \phi\right) - \phi$
$\dfrac{1}{2}mV'^2 = \dfrac{3}{2}\left(\dfrac{1}{2}mV^2\right) + \dfrac{\phi}{2}$
Since the work function $\phi > 0$, it follows that
$\dfrac{1}{2}mV'^2 > \dfrac{3}{2}\left(\dfrac{1}{2}mV^2\right)$
$V'^2 > 1.5 V^2$
$V' > V(1.5)^{\dfrac{1}{2}}$
Thus, the maximum speed of emitted photoelectrons will be greater than $V(1.5)^{\dfrac{1}{2}}$.
Hence, the correct answer is option: 'B'.