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: Use energies in units of hc/lambda
Let $E_1 = \dfrac{hc}{\lambda} = 1$ unit. Then the photon energy for $1.5\lambda$ is $\dfrac{2}{3}$ unit.

Step 2: Equations
$K_1 = 1 - \phi$ and $K_2 = \dfrac{2}{3} - \phi$, where $K_2 = \dfrac{K_1}{4}$.

Step 3: Solve
$\dfrac{1 - \phi}{4} = \dfrac{2}{3} - \phi$, so $1 - \phi = \dfrac{8}{3} - 4\phi$ and $3\phi = \dfrac{5}{3}$, giving $\phi = \dfrac{5}{9}$ unit.

Step 4: Threshold
Since $\phi = \dfrac{hc}{\lambda_0} = \dfrac{5}{9}\cdot\dfrac{hc}{\lambda}$, we get $\lambda_0 = \dfrac{9\lambda}{5}$.

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