A small object at rest, absorbs a light pulse of power 20 mW and duration 300 ns. Assuming speed of light as 3 × 10⁸ m/s, the momentum of the object becomes equal to: (a) 1 × 10⁻¹⁷ kg m/s (b) 0.5 × 10⁻¹⁷ kg m/s (c) 3 × 10⁻¹⁷ kg m/s (d) 2 × 10⁻¹⁷ kg m/s

Answer: (d). Energy absorbed E = Pt = 20×10⁻³× 300×10⁻⁹ = 6×10⁻⁹ J On absorption p = E/c = 6×10⁻⁹3×10⁸ = 2×10⁻¹⁷ kg m/s So the answer is (d) 2×10⁻¹⁷ kg m/s.

A small object at rest, absorbs a light pulse of power 20 mW and… | JEE Main 2023 (30 Jan, Shift 1)

Q17JEE Main 2023 · 30 JAN · SHIFT 1Easy

Photons: Energy, Momentum & Radiation Pressure

A small object at rest, absorbs a light pulse of power $20\text{ mW}$ and duration $300\text{ ns}$ . Assuming speed of light as $3 \times 10^{8}\text{ m/s}$ , the momentum of the object becomes equal to:

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Answer: (d)

Energy absorbed $E = Pt = 20\times10^{-3}\times 300\times10^{-9} = 6\times10^{-9}\text{ J}$
On absorption $p = \dfrac{E}{c} = \dfrac{6\times10^{-9}}{3\times10^{8}} = 2\times10^{-17}\text{ kg m/s}$
So the answer is (d) $2\times10^{-17}\text{ kg m/s}$ .

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