Photoelectric Effects and deBroglie Equation: Practice Problem & Solution
The momentum of a photon of an electromagnetic radiation is $3.3 \times 10^{-29} kg ms^{-1}$. What is the frequency of the associated waves? $[h = 6.6 \times 10^{-34} Js ; c = 3 \times 10^8 ms^{-1}]$ (1990)
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:
Momentum $p = \frac{h\nu}{c}$. Frequency $\nu = \frac{pc}{h} = \frac{3.3 \times 10^{-29} \times 3 \times 10^8}{6.6 \times 10^{-34}} = \frac{9.9 \times 10^{-21}}{6.6 \times 10^{-34}} = 1.5 \times 10^{13} Hz$.
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