Question

A 500 nm photon is incident normally on a perfectly reflecting surface and is reflected. The value of momentum transferred to the surface is:

• A. \( 3.87 \times 10^{-43} \, \text{kg} \, \text{ms}^{-1} \)
• B. \( 2.5 \times 10^{-30} \, \text{kg} \, \text{ms}^{-1} \)
• C. \( 2.65 \times 10^{-27} \, \text{kg} \, \text{ms}^{-1} \)
• D. \( 1.33 \times 10^{-27} \, \text{kg} \, \text{ms}^{-1} \)

Answer 1

The Correct Option is C

To find the momentum transferred to a perfectly reflecting surface by an incident photon, we can use the concept of photon momentum and its change upon reflection. Let's solve it step by step:

1. Firstly, find the momentum of the photon. The momentum \( p \) of a photon is given by the equation: \(p = \frac{h}{\lambda}\) where:
   • \( h \) is Planck's constant, \( 6.626 \times 10^{-34} \, \text{Js} \).
   • \( \lambda \) is the wavelength of the photon, given as 500 nm \( = 500 \times 10^{-9} \, \text{m} \).
2. Substitute the values into the momentum formula:

\(p_{\text{initial}} = \frac{6.626 \times 10^{-34}}{500 \times 10^{-9}}\)

1. Simplifying, we get:

\(p_{\text{initial}} = 1.3252 \times 10^{-27} \, \text{kg} \, \text{ms}^{-1}\)

1. For a perfectly reflecting surface, the photon rebounds with an equal and opposite momentum, and thus the change in momentum \( \Delta p \) is:

\(\Delta p = p_{\text{final}} - (-p_{\text{initial}}) = 2p_{\text{initial}}\)

\(\Delta p = 2 \times 1.3252 \times 10^{-27}\)

\(\Delta p = 2.65 \times 10^{-27} \, \text{kg} \, \text{ms}^{-1}\)

Therefore, the value of momentum transferred to the surface is \( 2.65 \times 10^{-27} \, \text{kg} \, \text{ms}^{-1} \), which corresponds to the correct answer.