Question

If the total energy transferred to a surface in time \( t \) is \( 6.48 \times 10^5 \, \text{J} \), then the magnitude of the total momentum delivered to this surface for complete absorption will be :

• A. \( 2.16 \times 10^{-3} \, \text{kg m/s} \)
• B. \( 2.46 \times 10^{-3} \, \text{kg m/s} \)
• C. \( 1.58 \times 10^{-3} \, \text{kg m/s} \)
• D. \( 4.32 \times 10^{-3} \, \text{kg m/s} \)

Answer 1

The Correct Option is A

The momentum (\( p \)) of electromagnetic radiation is related to its energy (\( E \)) by the equation:

\[ p = \frac{E}{c}, \]

where \( p \) is momentum, \( E \) is energy, and \( c \) is the speed of light (\( c \approx 3 \times 10^8 \, \text{m/s} \)).

Given an energy of:

\[ E = 6.48 \times 10^5 \, \text{J}. \]

The momentum can be calculated as:

\[ p = \frac{6.48 \times 10^5 \, \text{J}}{3 \times 10^8 \, \text{m/s}}. \]

This calculation yields:

\[ p = \frac{6.48}{3} \times 10^{-3} = 2.16 \times 10^{-3} \, \text{kg m/s}. \]

Therefore, the total momentum delivered to the surface upon complete absorption is:

\[ 2.16 \times 10^{-3} \, \text{kg m/s}. \]