Question

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 ? $[h=6.6 \times 10^{-34}\, Js\, ;\, c = 3 \times 10^8\, ms^{-1}]$

• A. $1.5 \times 10^{-13}\, Hz $
• B. $ 7.5 \times 10^{-12}\, Hz $
• C. $6 \times 10^{3}\, Hz $
• D. $ 3 \times 10^{3}\, Hz $

Answer 1

The Correct Option is A

To find the frequency of the associated waves, we start by understanding the relationship between the momentum of a photon and its frequency. The momentum p of a photon is given by the equation:

p = \frac{h}{\lambda}

where h is Planck's constant and \lambda is the wavelength of the photon. Additionally, the speed of light c is related to the frequency \nu and the wavelength \lambda by the equation:

c = \lambda \nu

From these two equations, we can express momentum in terms of frequency:

p = \frac{h \nu}{c}

We can rearrange this equation to solve for frequency \nu:

\nu = \frac{pc}{h}

Now, substitute the given values:

• p = 3.3 \times 10^{-29}\, \text{kg ms}^{-1}
• c = 3 \times 10^8\, \text{ms}^{-1}
• h = 6.6 \times 10^{-34}\, \text{Js}

Substituting these into the formula gives:

\nu = \frac{(3.3 \times 10^{-29}) \times (3 \times 10^8)}{6.6 \times 10^{-34}}

\nu = \frac{9.9 \times 10^{-21}}{6.6 \times 10^{-34}}

\nu = 1.5 \times 10^{13}\, \text{Hz}

Thus, the frequency of the associated wave is 1.5 \times 10^{13}\, \text{Hz}.

The correct answer is:

1.5 \times 10^{-13}\, \text{Hz}