Question

When radiation of wavelength '$\lambda$' is incident on a metallic surface, the stopping potential is $4.8 \text{ V}$. If the surface is illuminated with radiation of double the wavelength then the stopping potential becomes $1.6 \text{ V}$. The threshold wavelength for the surface is

• A. 2\lambda
• B. 4\lambda
• C. 6\lambda
• D. 8\lambda

Hint:

Use the equation $eV_s = hc(\frac{1}{\lambda} - \frac{1}{\lambda_0})$ for two cases and divide or substitute to eliminate constants like $hc$ and $e$.

Answer 1

The Correct Option is B

Step 1: Einstein's photoelectric equation is given by: \[ eV_0 = \frac{hc}{\lambda} - \phi_0 \] where $V_0$ is the stopping potential, $\lambda$ is the incident wavelength, and $\phi_0 = \frac{hc}{\lambda_0}$ is the work function of the metal ($ \lambda_0 $ is the threshold wavelength).
Step 2: For the first case, wavelength is $\lambda$ and stopping potential is $4.8 \text{ V}$: \[ 4.8 e = \frac{hc}{\lambda} - \phi_0 \quad \text{---(i)} \]
Step 3: For the second case, wavelength is $2\lambda$ and stopping potential is $1.6 \text{ V}$: \[ 1.6 e = \frac{hc}{2\lambda} - \phi_0 \quad \text{---(ii)} \]
Step 4: Multiplying equation (ii) by $3$: \[ 4.8 e = 3 \left( \frac{hc}{2\lambda} - \phi_0 \right) = \frac{3hc}{2\lambda} - 3\phi_0 \]
Step 5: Setting the expressions for $4.8 e$ from Step 2 and Step 4 equal to each other: \[ \frac{hc}{\lambda} - \phi_0 = \frac{3hc}{2\lambda} - 3\phi_0 \] \[ 2\phi_0 = \frac{3hc}{2\lambda} - \frac{hc}{\lambda} = \frac{hc}{2\lambda} \] \[ \phi_0 = \frac{hc}{4\lambda} \]
Step 6: Since $\phi_0 = \frac{hc}{\lambda_0}$: \[ \frac{hc}{\lambda_0} = \frac{hc}{4\lambda} \implies \lambda_0 = 4\lambda \]