Photoelectric Effects and deBroglie Equation: Practice Problem & Solution
15. The threshold frequency for a photosensitive metal is $3.3 \times 10^{14} Hz$. If light of frequency $8.2 \times 10^{14} Hz$ is incident on this metal, the cut-off voltage for the photoelectric emission is nearly: (2011 Mains)
Solution Explained:
To solve this problem, we apply the core principles of Photoelectric Effects and deBroglie Equation. Understanding the underlying formula is key to arriving at the correct answer below:
From photoelectric equation, $eV_0 = h(\nu - \nu_0)$. Thus, $V_0 = \frac{h(\nu - \nu_0)}{e} = \frac{6.6 \times 10^{-34} \times (8.2 - 3.3) \times 10^{14}}{1.6 \times 10^{-19}} = \frac{6.6 \times 4.9 \times 10^{-20}}{1.6 \times 10^{-19}} \approx 2 V$.
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