Question

In a gamma decay process, the internal energy of the nucleus of mass \(M\) decreases, a gamma photon of energy \(E\) and linear momentum \(\dfrac{E}{c}\) is emitted, and the nucleus recoils. The decrease of internal energy is:

• A. \(E\)
• B. \(E + \dfrac{E^2}{2Mc^2}\)
• C. \(E - \dfrac{E^2}{2Mc^2}\)
• D. \(\dfrac{E^2}{2Mc^2}\)

Hint:

Momentum starts at zero, so the nucleus recoils with momentum \(E/c\). Add the photon energy and the recoil kinetic energy \(p^2/2M\).

Answer 1

The Correct Option is B

Step 1: The nucleus is at rest before the decay, so the total momentum is zero. When the photon is emitted with momentum \(\dfrac{E}{c}\) in one direction, the nucleus must recoil with an equal and opposite momentum to keep the total momentum zero.

Step 2: So the recoil momentum of the nucleus is \(p = \dfrac{E}{c}\).

Step 3: The recoil kinetic energy of the nucleus (mass \(M\)) is\[K = \frac{p^2}{2M} = \frac{1}{2M}\left(\frac{E}{c}\right)^2 = \frac{E^2}{2Mc^2}.\]
Step 4: By conservation of energy, the internal energy lost by the nucleus is shared between the photon energy and the recoil kinetic energy of the nucleus.\[\Delta U = E + K = E + \frac{E^2}{2Mc^2}.\]This matches option (B).\[\boxed{\Delta U = E + \frac{E^2}{2Mc^2}}\]