Wavelength and Work Function from Photoelectric Stopping Potentials

To find the wavelength of the first light source and the work function of the metal, we use Einstein's photoelectric equation:

$e V_s = \frac{hc}{\lambda} - \Phi$

where:

• $V_s$ is the stopping potential,
• $\lambda$ is the wavelength of the incident light,
• $\Phi$ is the work function of the metal,
• $hc/e = 1.24 \times 10^{-6} \text{ J m C}^{-1} = 1.24 \times 10^{-6} \text{ eV m}$.

Note that the intensity of light affects only the photoelectric current and has no effect on the stopping potential.

Step 1: Setting up the equations for both cases

Case 1:
With stopping potential $V_1 = 6.0 \text{ V}$ and wavelength $\lambda_1 = \lambda$: $6.0 \text{ eV} = \frac{hc}{\lambda} - \Phi \quad \text{--- (1)}$

Case 2:
With stopping potential $V_2 = 0.6 \text{ V}$ and wavelength $\lambda_2 = 4\lambda$: $0.6 \text{ eV} = \frac{hc}{4\lambda} - \Phi \quad \text{--- (2)}$

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Step 2: Calculating the work function ($\Phi$)

Let $E = \frac{hc}{\lambda}$ be the photon energy of the first source in $\text{eV}$.

From Equation (1): $E = 6.0 + \Phi$

From Equation (2): $\frac{E}{4} = 0.6 + \Phi \implies E = 2.4 + 4\Phi$

Equating the two expressions for $E$: $6.0 + \Phi = 2.4 + 4\Phi$ $3\Phi = 6.0 - 2.4 = 3.6$ $\Phi = 1.20 \text{ eV}$

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Step 3: Calculating the wavelength of the first light source ($\lambda$)

Substitute $\Phi = 1.20 \text{ eV}$ back into the equation for $E$: $E = 6.0 + 1.20 = 7.20 \text{ eV}$

Since $E = \frac{hc}{e \lambda}$: $\lambda = \frac{hc / e}{E}$

Substituting the given value of $\frac{hc}{e} = 1.24 \times 10^{-6} \text{ eV m}$: $\lambda = \frac{1.24 \times 10^{-6}}{7.20} \approx 1.722 \times 10^{-7} \text{ m}$

Thus, the wavelength of the first light source is $1.72 \times 10^{-7} \text{ m}$ and the work function of the metal is $1.20 \text{ eV}$.

Correct Option: (A)