A Fe-57 nucleus is at rest and in its first excited state, 14.4 keV above the ground state (14.4 × 103 eV, where 1 eV = 1.6 ×
10−19 J). The nucleus then decays to the ground state with the emission of a gamma ray (a high-energy photon).
(a) What is the recoil speed of the nucleus?
(b) Calculate the slight difference in eV between the gammaray energy and the 14.4 keV difference between the initial
and final nuclear states.
(c) The “Mössbauer effect” is the name given to a related phenomenon discovered by Rudolf Mössbauer in 1957, for
which he received the 1961 Nobel Prize for physics. If the Fe-57 nucleus is in a solid block of iron, occasionally
when the nucleus emits a gamma ray the entire solid recoils as one object. This can happen due to the fact that
neighboring atoms and nuclei are connected by the electric interatomic force. In this case, repeat the calculation of
part (b) and compare with your previous result. Explain briefly.