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: When an atom emits a photon, momentum conservation gives the atom a recoil momentum equal in magnitude to the photon momentum. A photon of wavelength \(\lambda\) carries momentum \[p = \frac{h}{\lambda}.\]

Step 2: Write both recoil momenta: \[p_A = \frac{h}{\lambda_A} = \frac{h}{1500\ \text{nm}}, \qquad p_B = \frac{h}{\lambda_B} = \frac{h}{500\ \text{nm}}.\]

Step 3: Take the ratio; the constant \(h\) cancels: \[\frac{p_A}{p_B} = \frac{\lambda_B}{\lambda_A} = \frac{500}{1500} = \frac{1}{3}.\]

Step 4: Hence \(p_A:p_B = 1:3\).\[\boxed{p_A:p_B = 1:3}\]