Work Function and Stopping Potential in Photoelectric Effect

To determine the value of $\alpha$, we apply Einstein's photoelectric equation:

$E = \Phi + K_{\text{max}}$

where:

• $E$ is the energy of the incident photon,
• $\Phi$ is the work function of the metal surface,
• $K_{\text{max}}$ is the maximum kinetic energy of the emitted photoelectrons.

Step 1: Calculate the energy of the incident photon ($E$)

The energy of a photon of wavelength $\lambda = 331\text{ nm} = 331 \times 10^{-9}\text{ m}$ is given by:

$E = \frac{hc}{\lambda}$

Substituting the given constants $h = 6.62 \times 10^{-34}\text{ J s}$ and $c = 3 \times 10^8\text{ m/s}$:

$E = \frac{6.62 \times 10^{-34} \times 3 \times 10^8}{331 \times 10^{-9}}$

$E = \frac{19.86 \times 10^{-26}}{331 \times 10^{-9}} = 0.06 \times 10^{-17}\text{ J} = 6 \times 10^{-19}\text{ J}$

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Step 2: Calculate the maximum kinetic energy ($K_{\text{max}}$)

The maximum kinetic energy of the photoelectrons is related to the stopping potential $V_0 = 0.2\text{ V}$ by:

$K_{\text{max}} = e V_0$

Substituting $e = 1.6 \times 10^{-19}\text{ C}$:

$K_{\text{max}} = (1.6 \times 10^{-19}\text{ C}) \times 0.2\text{ V} = 0.32 \times 10^{-19}\text{ J}$

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Step 3: Calculate the work function ($\Phi$)

Rearranging Einstein's equation to solve for the work function:

$\Phi = E - K_{\text{max}}$

$\Phi = 6 \times 10^{-19}\text{ J} - 0.32 \times 10^{-19}\text{ J} = 5.68 \times 10^{-19}\text{ J}$

Equating this to the given expression $\Phi = \alpha \times 10^{-19}\text{ J}$:

$\alpha = 5.68$

Thus, the correct option is C.