Question

Following a nuclear reaction, a nucleus of aluminum is at rest in an excited state represented by 2713Al* , as shown above left. The excited nucleus returns to the ground state 2713Al by emitting a gamma ray photon of energy 1.02 MeV, as shown above right. The aluminum nucleus in the ground state has a mass of 4.48 x 10–26 kg.

(a) Calculate the wavelength of the emitted photon in meters.

(b) Calculate the momentum of the emitted photon in kg-m/*.

(c) Calculate the speed of the recoiling nucleus in m/*.

(d) Calculate the kinetic energy of the recoiling nucleus in joules.

Answer 1

(a) E = hc / λ … 1.02 x 106 eV = 1240 / λ … λ = 1.2 x 10–3 nm

(b) p = h / λ … 6.63 x 10–34 / 1.2x10–12 m … = 5.43 x 10–22 kg-m/*

(c) From conservation of momentum, the momentum before is zero so the momentum after is also zero. To conserve momentum after, the momentum of the photon must be equal and opposite to the momentum of the Al nucleus. pphoton = pnucleus = malval … 5.43 x 10–22 = 4.48 x 10–26 val … val = 1.21 x 104 m/*

(d) K = mv2 = (4.48 x 10–26)(1.21 x 104)2 = 3.28 x 10–18 J