Question

If the total momentum delivered to a surface by an EM wave is \(3 \times 10^{-4}\) kg m/s, then the total energy transferred to this surface is:

• A. \( 3 \times 10^4 \, {J} \)
• B. \( 4.5 \times 10^4 \, {J} \)
• C. \( 6 \times 10^4 \, {J} \)
• D. \( 2 \times 10^4 \, {J} \)
• E. \( 9 \times 10^4 \, {J} \)

Hint:

The energy transferred by an EM wave is the product of the momentum and the speed of light.

Answer 1

The Correct Option is A

The energy transferred by an EM wave is related to the momentum \( p \) and the speed of light \( c \) by the formula: \[ E = p \times c \] Where: - \( p \) is the momentum delivered to the surface,
- \( c \) is the speed of light in vacuum, \( c = 3 \times 10^8 \, {m/s} \).
Given: - \( p = 3 \times 10^{-4} \, {kg m/s} \),
- \( c = 3 \times 10^8 \, {m/s} \).
Thus, the energy \( E \) transferred is: \[ E = 3 \times 10^{-4} \times 3 \times 10^8 = 9 \times 10^4 \, {J} \] Therefore, the total energy transferred to the surface is \( 3 \times 10^4 \, {J} \).