Light with an energy flux of 500 kW/m2 falls for 5 minutes at normal incidence on a non-reflecting circular surface with a radius of 10 cm. The total momentum delivered to this surface has a magnitude of ______.
5π × 10-3 kg m s-1
Light Momentum Calculation on a Non-Reflecting Surface
This problem asks us to determine the total momentum delivered to a circular surface when light with a given energy flux falls on it for a specific duration. The surface is described as non-reflecting, which means it perfectly absorbs the incident light.
Key Concepts for Momentum Delivery
To solve this problem, we need to understand the relationship between light energy, intensity, and momentum transfer. When light falls on a surface, it exerts radiation pressure and delivers momentum. For a perfectly absorbing (non-reflecting) surface, the relationship between the total energy absorbed and the total momentum delivered is straightforward.
- Energy Flux (Intensity, I): This is the power per unit area carried by the light. It is given in kW/m2.
- Total Energy (U): The total energy incident on and absorbed by the surface over a given time is the product of intensity, area, and time.
\[U = I \times A \times t\]
- Momentum of Light: For a photon, momentum \(p\) and energy \(E\) are related by \(p = E/c\), where \(c\) is the speed of light. For a perfectly absorbing surface, the total momentum delivered (\(\Delta p\)) is the total energy absorbed (\(U\)) divided by the speed of light (\(c\)).
\[\Delta p = \frac{U}{c}\]
Step-by-Step Momentum Calculation
Let's break down the calculation into clear steps:
1. Convert Given Values to SI Units
It's crucial to work with consistent SI units for all quantities.
- Energy Flux (Intensity), I:
Given: \(I = 500 \, \text{kW/m}^2\)
Convert to W/m2: \(I = 500 \times 10^3 \, \text{W/m}^2 = 5.0 \times 10^5 \, \text{W/m}^2\) - Time, t:
Given: \(t = 5 \, \text{minutes}\)
Convert to seconds: \(t = 5 \times 60 \, \text{s} = 300 \, \text{s}\) - Radius of Circular Surface, r:
Given: \(r = 10 \, \text{cm}\)
Convert to meters: \(r = 10 \times 10^{-2} \, \text{m} = 0.1 \, \text{m}\) - Speed of Light, c:
This is a fundamental constant: \(c = 3.0 \times 10^8 \, \text{m/s}\)
2. Calculate the Area of the Circular Surface
The area (\(A\)) of a circular surface is given by the formula \(A = \pi r^2\).
Using the radius in meters:
\[A = \pi (0.1 \, \text{m})^2\] \[A = \pi (0.01) \, \text{m}^2\] \[A = 0.01 \pi \, \text{m}^2\]
3. Calculate the Total Energy Delivered to the Surface
The total energy (\(U\)) delivered to the surface is the product of the energy flux (intensity), the area, and the time duration.
\[U = I \times A \times t\] \[U = (5.0 \times 10^5 \, \text{W/m}^2) \times (0.01 \pi \, \text{m}^2) \times (300 \, \text{s})\] \[U = (5.0 \times 10^5) \times (\pi \times 10^{-2}) \times (3 \times 10^2) \, \text{J}\]
Now, let's group the numerical terms and powers of 10:
\[U = (5.0 \times 3) \times \pi \times (10^5 \times 10^{-2} \times 10^2) \, \text{J}\] \[U = 15 \pi \times (10^{5 - 2 + 2}) \, \text{J}\] \[U = 15 \pi \times 10^5 \, \text{J}\]4. Calculate the Total Momentum Delivered
Since the surface is non-reflecting (perfectly absorbing), the total momentum delivered (\(\Delta p\)) is the total energy absorbed divided by the speed of light.
\[\Delta p = \frac{U}{c}\] \[\Delta p = \frac{15 \pi \times 10^5 \, \text{J}}{3.0 \times 10^8 \, \text{m/s}}\]
Divide the numerical terms and powers of 10 separately:
\[\Delta p = \left(\frac{15}{3}\right) \pi \times \left(\frac{10^5}{10^8}\right) \, \text{kg m s}^{-1}\] \[\Delta p = 5 \pi \times 10^{5-8} \, \text{kg m s}^{-1}\] \[\Delta p = 5 \pi \times 10^{-3} \, \text{kg m s}^{-1}\]Summary of Result
The total momentum delivered to the non-reflecting circular surface is \(5 \pi \times 10^{-3} \, \text{kg m s}^{-1}\).
| Quantity | Symbol | Value | Units |
|---|---|---|---|
| Energy Flux | \(I\) | \(500 \times 10^3\) | W/m2 |
| Time | \(t\) | \(300\) | s |
| Radius | \(r\) | \(0.1\) | m |
| Area | \(A\) | \(0.01 \pi\) | m2 |
| Speed of Light | \(c\) | \(3 \times 10^8\) | m/s |
| Total Energy Delivered | \(U\) | \(15 \pi \times 10^5\) | J |
| Total Momentum Delivered | \(\Delta p\) | \(5 \pi \times 10^{-3}\) | kg m s-1 |
This result matches one of the provided options, confirming our calculation for the total momentum delivered by the light.
Important Questions from Momentum and Energy
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