Question

When a certain metallic surface is illuminated with monochromatic light of wavelength $\lambda$, the stopping potential for photoelectric effect is $3V_0$. If the same surface is illuminated with light of wavelength $2\lambda$, the stopping potential is found as $V_0$. The threshold wavelength for this surface is

• A. $\dfrac{\lambda}{4}$
• B. $4\lambda$
• C. $6\lambda$
• D. $\dfrac{\lambda}{6}$

Hint:

Threshold wavelength depends only on the material, not on intensity or stopping potential.

Answer 1

The Correct Option is B

Step 1: Write photoelectric equation.
\[ eV = \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_0} \] Step 2: Write equations for given wavelengths.
For wavelength $\lambda$: \[ e(3V_0) = \dfrac{hc}{\lambda} - \dfrac{hc}{\lambda_0} \] For wavelength $2\lambda$: \[ e(V_0) = \dfrac{hc}{2\lambda} - \dfrac{hc}{\lambda_0} \] Step 3: Subtract the equations.
\[ 2eV_0 = \dfrac{hc}{\lambda} - \dfrac{hc}{2\lambda} = \dfrac{hc}{2\lambda} \] Step 4: Find $V_0$.
\[ eV_0 = \dfrac{hc}{4\lambda} \] Step 5: Substitute back to find threshold wavelength.
From second equation: \[ \dfrac{hc}{\lambda_0} = \dfrac{hc}{2\lambda} - \dfrac{hc}{4\lambda} = \dfrac{hc}{4\lambda} \] \[ \lambda_0 = 4\lambda \]