Q138 / 150 · MHT CET 2026 (16 April Shift 1) · Physics
For a photosensitive material, work function is '$W_0$' and stopping potential is 'V'. The wavelength of the incident radiation is (h=Planck's constant, c = velocity of light, e = electronic charge)
- A$\dfrac{\text{hc}}{(\text{W}_0 - \text{eV})}$
- Correct answerB$\dfrac{\text{hc}}{(\text{W}_0 + \text{eV})}$
- C$\text{hc}(\text{W}_0 - \text{eV})$
- D$\text{hc}(\text{W}_0 + \text{eV})$
Solution
According to Einstein's photoelectric equation:
$E = W_0 + K_{\text{max}}$
where $E$ is the energy of the incident radiation, $W_0$ is the work function, and $K_{\text{max}}$ is the maximum kinetic energy of the emitted photoelectrons.
The maximum kinetic energy is related to the stopping potential $V$ by:
$K_{\text{max}} = eV$
The energy of the incident radiation is given by:
$E = \dfrac{hc}{\lambda}$
Equating the expressions for $E$:
$\dfrac{hc}{\lambda} = W_0 + eV$
Solving for $\lambda$:
$\lambda = \dfrac{hc}{W_0 + eV}$
Answer: $\dfrac{\text{hc}}{(\text{W}_0 + \text{eV})}$
$E = W_0 + K_{\text{max}}$
where $E$ is the energy of the incident radiation, $W_0$ is the work function, and $K_{\text{max}}$ is the maximum kinetic energy of the emitted photoelectrons.
The maximum kinetic energy is related to the stopping potential $V$ by:
$K_{\text{max}} = eV$
The energy of the incident radiation is given by:
$E = \dfrac{hc}{\lambda}$
Equating the expressions for $E$:
$\dfrac{hc}{\lambda} = W_0 + eV$
Solving for $\lambda$:
$\lambda = \dfrac{hc}{W_0 + eV}$
Answer: $\dfrac{\text{hc}}{(\text{W}_0 + \text{eV})}$
Hence, the correct answer is option: 'B'.