Question

When light of wavelength '\(λ\)' is incident on a photosensitive surface, the stopping potential is 'V'. When a light of wavelength \(1.5λ\) is incident on the same surface, the stopping potential is '\(\frac{V}{4}\)'. Threshold wavelength for the surface is

• A. \(\frac{6}{5}λ\)
• B. \(\frac{7.5}{4}λ\)
• C. \(\frac{7.5}{9}λ\)
• D. \(\frac{9}{5}λ\)

Hint:

Write Einstein equations for both wavelengths and eliminate the work function to find the threshold wavelength.

Answer 1

The Correct Option is D

Step 1: Understand the concept
Einstein's equation gives \(eV_0 = \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_0}\), where \(\lambda_0\) is the threshold wavelength.

Step 2: Write the two cases
Case 1: \(eV = \dfrac{hc}{\lambda} - \phi\). Case 2: \(\dfrac{eV}{4} = \dfrac{hc}{1.5\lambda} - \phi\), where \(\phi = \dfrac{hc}{\lambda_0}\).

Step 3: Eliminate \(\phi\)
Subtract: \(\dfrac{3eV}{4} = \dfrac{hc}{\lambda}\left(1 - \dfrac{2}{3}\right) = \dfrac{hc}{3\lambda}\), so \(eV = \dfrac{4hc}{9\lambda}\).

Step 4: Find the threshold
\(\phi = \dfrac{hc}{\lambda} - eV = \dfrac{hc}{\lambda}\left(1 - \dfrac{4}{9}\right) = \dfrac{5hc}{9\lambda}\). So
\[ \lambda_0 = \frac{hc}{\phi} = \frac{9}{5}\lambda \]
Option (D).

Final Answer:
The threshold wavelength is 9 lambda / 5. This is option (D). \[ \boxed{\text{(D) }\frac{9}{5}\lambda} \]