Question

A pulse of radiation is absorbed by an object initially at rest for \(10^{-4}\, s\). If the power of the pulse is \(9 \times 10^{-3}\, W\), then the total momentum of the object received is. (Speed of light in vacuum \(c = 3 \times 10^8\, m\,s^{-1}\))

• A. \(3 \times 10^{8}\, kg\, m\, s^{-1}\)
• B. \(3 \times 10^{13}\, kg\, m\, s^{-1}\)
• C. \(3 \times 10^{-15}\, kg\, m\, s^{-1}\)
• D. \(3\, kg\, m\, s^{-1}\)

Hint:

For electromagnetic waves, momentum is always \(p = E/c\), and energy is \(E = Pt\).

Answer 1

The Correct Option is C

Step 1: Use the force exerted by absorbed radiation, instead of computing total energy and momentum together.
An object that fully absorbs radiation of power $P$ feels a steady force \[ F = \frac{P}{c} \] since each second the energy $P$ (and the momentum $P/c$ that goes with it) is delivered and absorbed.

Step 2: Compute this force.
\[ F = \frac{9 \times 10^{-3}}{3 \times 10^{8}} = 3 \times 10^{-11}\ \text{N} \]
Step 3: Get the momentum delivered as force times time, since the power is steady over the pulse.
\[ p = F \times t = 3 \times 10^{-11} \times 10^{-4} \]
Step 4: Work out the multiplication.
\[ p = 3 \times 10^{-15}\ \text{kg m s}^{-1} \]
Step 5: Why the other options are off.
$3 \times 10^8$ and $3 \times 10^{13}$ come from misplacing powers of ten; $3\ \text{kg m s}^{-1}$ would need vastly more energy than this weak, brief pulse carries.

Final Answer:
\[ \boxed{3 \times 10^{-15}\ \text{kg m s}^{-1}} \]