Question:
Find the frequency of light which ejects electrons from a metal surface fully stopped by a retarding potential of 3 V. The photoelectric effect begins in this metal at a frequency of \(6 \times 10^{15}\) Hz.
Find the frequency of light which ejects electrons from a metal surface fully stopped by a retarding potential of 3 V. The photoelectric effect begins in this metal at a frequency of \(6 \times 10^{15}\) Hz.
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In the photoelectric effect, $eV_s = h\nu - \phi = h(\nu - \nu_0)$. Solve for $\nu = eV_s/h + \nu_0$.
Updated On: Apr 20, 2026
- \(1.324 \times 10^{15}\) Hz
- \(2.295 \times 10^{16}\) Hz
- \(3.678 \times 10^{18}\) Hz
- \(2.7 \times 10^{14}\) Hz
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Einstein's photoelectric equation: \(eV_s = h(\nu - \nu_0)\), where \(\nu_0\) is the threshold frequency.
Step 2: Detailed Explanation:
\[ \nu = \frac{eV_s}{h} + \nu_0 = \frac{1.6\times10^{-19}\times3}{6.63\times10^{-34}} + 6\times10^{15} \] \[ = \frac{4.8\times10^{-19}}{6.63\times10^{-34}} + 6\times10^{15} = 7.23\times10^{14} + 6\times10^{14} \times 10 \] \[ = 7.23\times10^{14} + 60\times10^{14} = 13.23\times10^{14} = 1.323\times10^{15} \text{ Hz} \]
Step 3: Final Answer:
Required frequency \(\approx 1.324 \times 10^{15}\) Hz.
Einstein's photoelectric equation: \(eV_s = h(\nu - \nu_0)\), where \(\nu_0\) is the threshold frequency.
Step 2: Detailed Explanation:
\[ \nu = \frac{eV_s}{h} + \nu_0 = \frac{1.6\times10^{-19}\times3}{6.63\times10^{-34}} + 6\times10^{15} \] \[ = \frac{4.8\times10^{-19}}{6.63\times10^{-34}} + 6\times10^{15} = 7.23\times10^{14} + 6\times10^{14} \times 10 \] \[ = 7.23\times10^{14} + 60\times10^{14} = 13.23\times10^{14} = 1.323\times10^{15} \text{ Hz} \]
Step 3: Final Answer:
Required frequency \(\approx 1.324 \times 10^{15}\) Hz.
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