Photoelectric Effect Analysis

The photoelectric effect relates photon energy ($E_{photon}$) to the metal's work function ($\phi$) and the maximum kinetic energy ($KE_{max}$) of emitted electrons using Einstein's equation: $E_{photon} = \phi + KE_{max}$.

Expressing energy in terms of wavelength ($\lambda$) and stopping potential ($V_s$), we have $E_{photon} = \frac{hc}{\lambda}$ and $KE_{max} = eV_s$. The equation becomes $\frac{hc}{\lambda} = \phi + eV_s$.

Formulating Equations

We are given:

• Light 1: Wavelength = $\lambda_1$, Stopping Potential $V_{s1} = 3.2 \text{ V}$.
• Light 2: Wavelength = $\lambda_2 = 2\lambda_1$, Stopping Potential $V_{s2} = 0.7 \text{ V}$.

Applying the photoelectric equation for both cases:

1. For Light 1: $ \frac{hc}{\lambda_1} = \phi + e(3.2) \quad \quad (1) $
2. For Light 2: $ \frac{hc}{2\lambda_1} = \phi + e(0.7) \quad \quad (2) $

Solving for $\lambda_1$

Subtract equation (2) from equation (1) to eliminate the work function ($\phi$):

$ (\frac{hc}{\lambda_1} - \frac{hc}{2\lambda_1}) = (\phi + 3.2e) - (\phi + 0.7e) $

$ \frac{hc}{2\lambda_1} = (3.2 - 0.7)e $

$ \frac{hc}{2\lambda_1} = 2.5e $

Rearranging to solve for $\lambda_1$:

$ \lambda_1 = \frac{hc}{5e} $

Calculation

Substitute the given physical constants: $h = 6.63 \times 10^{-34} \text{ J.s}$, $c = 3 \times 10^8 \text{ m/s}$, and $e = 1.6 \times 10^{-19} \text{ C}$:

$ \lambda_1 = \frac{(6.63 \times 10^{-34} \text{ J.s}) \times (3 \times 10^8 \text{ m/s})}{5 \times (1.6 \times 10^{-19} \text{ C})} $

$ \lambda_1 = \frac{19.89 \times 10^{-26} \text{ J.m}}{8 \times 10^{-19} \text{ C}} $

$ \lambda_1 = 2.48625 \times 10^{-7} \text{ m} $

The calculation yields $\lambda_1 \approx 2.49 \times 10^{-7}$ m. The provided correct option is B.