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Physics Question on Photoelectric Effect

The momentum of a photon of an electromagnetic radiation is 3.3×10−29 kg ms−1.3.3 \times 10^{-29}\, kg\, ms^{-1}. What is the frequency of the associated waves ? [h=6.6×10−34 Js ; c=3×108 ms−1][h=6.6 \times 10^{-34}\, Js\, ;\, c = 3 \times 10^8\, ms^{-1}]

A

1.5×10−13 Hz1.5 \times 10^{-13}\, Hz

B

7.5×10−12 Hz7.5 \times 10^{-12}\, Hz

C

6×103 Hz6 \times 10^{3}\, Hz

D

3×103 Hz3 \times 10^{3}\, Hz

Answer

1.5×10−13 Hz1.5 \times 10^{-13}\, Hz

Explanation

Solution

Momentum of the photon =hυc= \frac{h\upsilon}{c}
⇒cυ=hp=λ\Rightarrow \frac{c}{\upsilon} =\frac{h}{p}=\lambda
υcλ=cph=3× 108×3.3×10−296.6×10−34\upsilon\frac{c}{\lambda} =\frac{cp}{h}=3 \times \, 10^8 \times \frac{3.3 \times 10^{-29}}{6.6 \times 10^{-34}}
=1.5×1013 Hz= 1.5 \times 10^{13}\, Hz
where, υ \upsilon = frequency of radiation