Question

The recoil momentum of an atom is \(p_A\) when it emits an infrared photon of wavelength 1500 nm, and \(p_B\) when it emits a photon of visible wavelength 500 nm. The ratio \(p_A/p_B\) is:

• A. \(1:1\)
• B. \(1:\sqrt{3}\)
• C. \(1:3\)
• D. \(3:2\)

Hint:

Photon momentum \(p = h/\lambda\), so recoil momentum is inversely proportional to wavelength: \(p_A/p_B = \lambda_B/\lambda_A\).

Answer 1

The Correct Option is C

Step 1: The recoil momentum of the atom is fixed by the photon it emits. Using the de Broglie / photon relation, momentum is inversely proportional to wavelength: $p \propto 1/\lambda$.

Step 2: Therefore the ratio of recoil momenta depends only on the ratio of wavelengths, inverted: \[\frac{p_A}{p_B} = \frac{1/\lambda_A}{1/\lambda_B} = \frac{\lambda_B}{\lambda_A}.\]

Step 3: Substitute $\lambda_A = 1500$ nm (infrared) and $\lambda_B = 500$ nm (visible): \[\frac{p_A}{p_B} = \frac{500}{1500} = \frac{1}{3}.\]

Step 4: The longer-wavelength infrared photon carries less momentum, so the atom recoils less when emitting it. The visible photon (three times shorter wavelength) gives three times the recoil. Hence the ratio is $1:3$.\[\boxed{p_A:p_B = 1:3}\]