Answer to Question #92220 in Atomic and Nuclear Physics for M M

Question

Each of the following nuclei emits a photon in a y transition between an excited state and the ground state. Given the energy of the photon, find the energy of the excited state and comment on the relationship between the nuclear recoil energy and the experimental uncertainty in the photon energy: (a) 320.08419 0.00042 keV in 51V; (b) 1475.786 f 0.005 keV in "'Cd; (c) 1274.545 &- 0.017 keV in 22Ne; (d) 3451.152 2 0.047 keV in 56Fe; (e) 884.54174 f 0.00074 keV in 19,1r.

Answer 1

The energy of the ground state for hydrogen-like atoms is

$$E_{gr}=-Z^2\cdot13.61\text{ MeV}.$$

According to law of conservation of energy, we can write

$$E_{ex}=E_{gr}+E_{rec}+E_\gamma=-Z^2\cdot13.61+E_{rec}+E_\gamma.$$

the recoil energy $E_{rec}$ corresponds with the experimental uncertainty of the energy of photon, and the difference between energies of the excited and ground levels will be greater than the photon energy. Also, the energy of the excited state must be less than the energy of the ground state (because at infinity the energy is zero). Thus, for the cases above

(a)

$$E_{ex}=-35.39929\text{ MeV}.$$

(b)

$$E_{ex}=-164.6795\text{ MeV}.$$

(c)

$$E_{ex}=-6.585965\text{ MeV}.$$

(d)

$$E_{ex}=-42.67751\text{ MeV}.$$

(e)

$$E_{ex}=-501.7182\text{ MeV}.$$

As we can see, the experimental uncertainty, or recoil energy, does not significantly affect the value of the energy of the excited state.