Question

On a surface, if photon of wavelength $\lambda$ is incident, the stopping potential is $3.2$ V. If the wavelength incident is $2\lambda$, stopping potential is $0.7$ V. Find $\lambda$.

• A. $4.96 \times 10^{-7}$ m
• B. $3.62 \times 10^{-7}$ m
• C. $7.24 \times 10^{-7}$ m
• D. $2.48 \times 10^{-7}$ m

Hint:

Subtracting photoelectric equations removes work function and simplifies calculations.

Answer 1

The Correct Option is D

Step 1: Photoelectric equation.
\[ eV = \dfrac{hc}{\lambda} - \phi \]
Step 2: Writing equations for both cases.
For wavelength $\lambda$:
\[ e(3.2) = \dfrac{hc}{\lambda} - \phi \quad \text{(1)} \]
For wavelength $2\lambda$:
\[ e(0.7) = \dfrac{hc}{2\lambda} - \phi \quad \text{(2)} \]
Step 3: Subtracting equation (2) from (1).
\[ e(2.5) = \dfrac{hc}{2\lambda} \]
Step 4: Solving for $\lambda$.
\[ \lambda = \dfrac{hc}{5e} \]
Using $hc = 12400$ eV\AA :
\[ \lambda = \dfrac{12400}{5} \,\text{\AA} = 2480\,\text{\AA} \]
\[ \lambda = 2.48 \times 10^{-7}\,\text{m} \]