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.

• A. \(1.324 \times 10^{15}\) Hz
• B. \(2.295 \times 10^{16}\) Hz
• C. \(3.678 \times 10^{18}\) Hz
• D. \(2.7 \times 10^{14}\) Hz

Hint:

In the photoelectric effect, $eV_s = h\nu - \phi = h(\nu - \nu_0)$. Solve for $\nu = eV_s/h + \nu_0$.

Answer 1

The Correct Option is A

Step 1: Underlying Concept:
Einstein's photoelectric equation: \(eV_s = h(\nu - \nu_0)\), where \(\nu_0\) is the threshold frequency.
Step 2: 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: Conclusion:
Required frequency \(\approx 1.324 \times 10^{15}\) Hz.