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

Answer: (A)

$2.16 \times 10^{-3} \, \text{kg m/s}$

Answer 2

The relationship between energy $E$ and momentum $p$ for electromagnetic radiation can be expressed as:

$p = \frac{E}{c},$

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

Given:

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

Substituting the values into the momentum formula:

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

Calculating:

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

Thus, the magnitude of the total momentum delivered to this surface for complete absorption is:

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