Question

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. \(\frac{\text{hc}}{(\text{W}_0-\text{eV})}\)
• B. \(\frac{\text{hc}}{(\text{W}_0+\text{eV})}\)
• C. \(\text{hc}(\text{W}_0-\text{eV})\)
• D. \(\text{hc}(\text{W}_0+\text{eV})\)

Hint:

Einstein's equation: photon energy equals work function plus maximum kinetic energy.

Answer 1

The Correct Option is B

Step 1: Energy sharing:
Part of the photon energy frees the electron ($W_0$); the rest is its kinetic energy, stopped by the potential $V$, so equal to $eV$.

Step 2: Substitute:
$\dfrac{hc}{\lambda} = W_0 + eV$.

Step 3: Result:
$\lambda = \dfrac{hc}{W_0 + eV}$, option (B).

Final Answer:
The wavelength is hc / (W0 + eV). \[ \boxed{\text{(B) }\dfrac{hc}{W_0+eV}} \]