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

We are given a 500 nm photon that is incident normally on a perfectly reflecting surface. The goal is to find the value of momentum transferred to the surface.

Step 1: Calculate the energy of the photon.
The energy \( E \) of the photon is given by the equation:
\[ E = \frac{hc}{\lambda} \] where:
- \( h = 6.626 \times 10^{-34} \, \text{J·s} \) (Planck's constant)
- \( c = 3 \times 10^8 \, \text{m/s} \) (speed of light)
- \( \lambda = 500 \times 10^{-9} \, \text{m} \) (wavelength of the photon)
Substituting the values:
\[ E = \frac{(6.626 \times 10^{-34}) \times (3 \times 10^8)}{500 \times 10^{-9}} = 3.9756 \times 10^{-19} \, \text{J} \]
Step 2: Calculate the momentum of the photon.
The momentum \( p \) of a photon is given by:
\[ p = \frac{E}{c} \] Substituting the value of \( E \):
\[ p = \frac{3.9756 \times 10^{-19}}{3 \times 10^8} = 1.325 \times 10^{-27} \, \text{kg·m/s} \]
Step 3: Calculate the momentum transferred to the surface.
Since the photon is reflected, the momentum transferred to the surface is twice the photon's momentum:
\[ p_{\text{transfer}} = 2p = 2 \times 1.325 \times 10^{-27} = 2.65 \times 10^{-27} \, \text{kg·m/s} \] Final Answer: \( 2.65 \times 10^{-27} \, \text{kg} \, \text{ms}^{-1} \)