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 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:

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

Answer 2

The problem asks for the magnitude of the total momentum delivered to a surface, given the total energy transferred to it, assuming the energy is completely absorbed.

Concept Used:

This problem is based on the dual nature of electromagnetic radiation, specifically its particle nature. According to the quantum theory of light, electromagnetic radiation consists of discrete packets of energy called photons. Each photon possesses both energy and momentum.

The relationship between the energy (\(E\)) of a photon and its momentum (\(p\)) is given by the equation:

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

where \(c\) is the speed of light in a vacuum (\(c \approx 3 \times 10^8 \, \text{m/s}\)). This relationship holds true not just for a single photon but for the total energy and total momentum of a beam of radiation.

When radiation is "completely absorbed" by a surface, it means the entire momentum of the incident radiation is transferred to the surface. Therefore, the momentum delivered to the surface is equal to the total momentum of the incident radiation.

Step-by-Step Solution:

Step 1: List the given information and the required constant.

The total energy transferred to the surface is given as:

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

The speed of light in a vacuum is a fundamental constant:

\[ c = 3 \times 10^8 \, \text{m/s} \]

The problem states that the absorption is complete, so the momentum delivered to the surface (\(P\)) is equal to the total momentum of the incident energy.

Step 2: Apply the energy-momentum relation.

The formula that connects the total energy (\(E\)) of the radiation and its total momentum (\(P\)) is:

\[ P = \frac{E}{c} \]

Step 3: Substitute the given values into the formula to calculate the momentum.

We substitute the values of \(E\) and \(c\) into the equation:

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

Final Computation & Result:

Now, we perform the calculation:

First, divide the numerical parts:

\[ \frac{6.48}{3} = 2.16 \]

Next, divide the powers of ten:

\[ \frac{10^5}{10^8} = 10^{5-8} = 10^{-3} \]

Combining these results, we get the magnitude of the total momentum:

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

The unit is kg·m/s, as \( \text{J}/(\text{m/s}) = (\text{kg} \cdot \text{m}^2/\text{s}^2) / (\text{m/s}) = \text{kg} \cdot \text{m/s} \).

Therefore, the magnitude of the total momentum delivered to the surface is \( 2.16 \times 10^{-3} \, \text{kg} \cdot \text{m/s} \).