Light of wavelength '$λ$' which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with some velocity then stopping potential

Question

The difference between threshold wavelengths for two metal surfaces $ A $ and $ B $ having work functions $ \phi_A = 9 \, \text{eV} $ and $ \phi_B = 4.5 \, \text{eV} $ is: \[ \text{(Given: } h c = 1242 \, \text{eV nm)} \]

Answer 1

Step 1: Formula:
$V_0=\dfrac{hc}{e}\left(\dfrac1\lambda-\dfrac1{\lambda_0}\right)$.

Step 2: Change:
Smaller $\lambda$ makes $\dfrac1\lambda$ larger, so $V_0$ becomes larger.

Step 3: Option:
(B).

Final Answer:
Smaller wavelength gives bigger photon energy. \[ \boxed{B} \]

Light of wavelength '\(λ\)' which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with some velocity then stopping potential

Show Hint

The Correct Option is B

Solution and Explanation

Top Questions on Photoelectric Effect

Questions Asked in MHT CET exam