Photoelectric Effect Problem
Question: When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is 8.0 V. This potential drops to 0.8 V if another source—with wavelength five times that of the first one and intensity two‐thirds of the first—is used. The wavelength of the first source is \[ \lambda_1=\frac{1.24}{n}\times10^{-6}~\text{m}, \] and the work function of the metal is \(x\) eV. Determine the values of \(n\) and \(x\). (Take \(\frac{hc}{e} = 1.24 \times 10^{-6}\) J·m·C\(^{-1}\)).
Detailed Step-by-Step Explanation
Step 1: For a photoelectric effect, the maximum kinetic energy of the emitted electrons is given by: \[ K_{\text{max}} = hf - \phi, \] where \(\phi\) is the work function of the metal. When a stopping potential \(V_0\) is applied, we have: \[ eV_0 = K_{\text{max}} = hf - \phi. \]
Step 2: First Source
For the first source with stopping potential \(V_1 = 8.0~\text{V}\) and wavelength \(\lambda_1\), note that:
\[
e \times 8.0 = \frac{hc}{\lambda_1} - \phi.
\]
Using the given expression for wavelength:
\[
\lambda_1 = \frac{1.24}{n} \times 10^{-6}~\text{m},
\]
and using the provided value \(\frac{hc}{e} = 1.24 \times 10^{-6}\) (which effectively means \(\frac{hc}{\lambda_1}\) in volts when \(\lambda_1\) is in meters), we find:
\[
8.0 = \frac{1.24 \times 10^{-6}}{\frac{1.24}{n} \times 10^{-6}} - \phi = n - \phi.
\]
Thus,
\[
\phi = n - 8.0. \quad \text{(Equation 1)}
\]
Step 3: Second Source
The second source has a wavelength five times that of the first:
\[
\lambda_2 = 5\lambda_1 = 5\left(\frac{1.24}{n} \times 10^{-6}\right) = \frac{6.20}{n} \times 10^{-6}~\text{m}.
\]
For this source, the stopping potential is \(V_2 = 0.8~\text{V}\). Thus:
\[
0.8 = \frac{hc/e}{\lambda_2} - \phi = \frac{1.24 \times 10^{-6}}{\frac{6.20}{n} \times 10^{-6}} - \phi = \frac{n}{5} - \phi.
\]
Rearranging gives:
\[
\phi = \frac{n}{5} - 0.8. \quad \text{(Equation 2)}
\]
Step 4: Equate Equations (1) and (2) to solve for \(n\): \[ n - 8.0 = \frac{n}{5} - 0.8. \] Multiply both sides by 5: \[ 5n - 40 = n - 4. \] Rearranging: \[ 5n - n = 40 - 4 \quad \Rightarrow \quad 4n = 36 \quad \Rightarrow \quad n = 9. \]
Step 5: Now substitute \(n = 9\) back into Equation (1): \[ \phi = 9 - 8.0 = 1~\text{eV}. \]
Final Answer: The value of \(n\) is 9 and the work function \(\phi\) is 1 eV, which corresponds to option (2).
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