Subfield: Modern Physics


Question: b) The ${ }^{58} \mathrm{Ni}$ nucleus produced in the excited state discussed in the part a), deexcites into its ground state by emitting a gamma-photon in the direction of its motion. Consider this decay in the frame of reference in which ${ }^{58} \mathrm{Ni}$ is at rest to find the recoil energy of ${ }^{58} \mathrm{Ni}$ (i.e. kinetic energy which ${ }^{58} \mathrm{Ni}$ acquires after the emission of the photon). What is the photon energy in that system? What is the photon energy in the lab system of reference (i.e. what would be the energy of the photon measured in the detector which is positioned in the direction in which the ${ }^{58} \mathrm{Ni}$ nucleus moves)?

![](https://cdn.mathpix.com/cropped/2023_12_21_566d5d2f42cbf0437902g-1.jpg?height=714&width=1459&top_left_y=1813&top_left_x=296)

Solution:
For gamma-emission from the static nucleus, laws of conservation of energy and momentum give:
$$
\begin{array}{r}
E_{x}\left({ }^{58} \mathrm{Ni}\right)=E_{\gamma}+E_{\text {recoil }} \\
p_{\gamma}=p_{\text {recoil }}
\end{array}
\tag{36}
$$
Gamma-ray and recoiled nucleus have, of course, opposite directions. For gamma-ray (photon), energy and momentum are related as:
$$
E_{\gamma}=p_{\gamma} \cdot c
\tag{37}
$$
In part a) we have seen that the nucleus motion in this energy range is not relativistic, so we have:
$$
E_{\text {recoil }}=\frac{p_{\text {recii }}^{2}}{2 m\left({ }^{58} \mathrm{Ni}\right)}=\frac{p_{\gamma}^{2}}{2 m\left({ }^{58} \mathrm{Ni}\right)}=\frac{E_{\gamma}^{2}}{2 m\left({ }^{(58} \mathrm{Ni}\right) \cdot c^{2}}
\tag{38}
$$
Inserting this into law of energy conservation Eq. (36), we get:
$$
E_{x}\left({ }^{58} \mathrm{Ni}\right)=E_{\gamma}+E_{\text {recoil }}=E_{\gamma}+\frac{E_{\gamma}^{2}}{2 m\left({ }^{58} \mathrm{Ni}\right) \cdot c^{2}}
\tag{39}
$$
This reduces to the quadratic equation:
$$
E^2_{\gamma}+2m({}^{58}Ni)c^2\cdot E_{\gamma}+2m({}^{58}Ni)c^2 E_x({}^{58}Ni)=0
\tag{40}
$$

$$
\begin{aligned}
E_{\gamma} &= \frac{-2m({}^{58}Ni)c^2+\sqrt{4(m({}^{58}Ni)c^2)^2+8m({}^{58}Ni)c^2E_x({}^{58}Ni)}}{2} \\
&=\sqrt{(m({}^{58}Ni)c^2)^2+2m({}^{58}Ni)c^2E_x({}^{58}Ni)}-m({}^{58}Ni)c^2
\end{aligned}
\tag{41}
$$
Inserting numbers gives:
$$
E_{\gamma}=10.8633 \mathrm{MeV}
\tag{42}
$$
The equation (37) can also be reduced to an approximate equation before inserting numbers:
$$
E_{\gamma}=E_{x}\left(1-\frac{E_{x}}{2 m\left({ }^{58} \mathrm{Ni}\right) c^{2}}\right)=10.8633 \mathrm{MeV}
\tag{43}
$$
The recoil energy is now easily found as:
$$
E_{\text {recoil }}=E_{x}\left({ }^{58} \mathrm{Ni}\right)-E_{\gamma}=1.1 \mathrm{keV}
\tag{44}
$$
Due to the fact that nucleus emitting gamma-ray $\left({ }^{58} \mathrm{Ni}\right)$ is moving with the high velocity, the energy of gamma ray will be changed because of the Doppler effect. The relativistic Doppler effect (when source is moving towards observer/detector) is given with this formula:
$$
f_{\text {detector }}=f_{\gamma, \text { emitted }} \sqrt{\frac{1+\beta}{1-\beta}}
\tag{45}
$$
and since there is a simple relation between photon energy and frequency ( $E=h f)$, we get the similar expression for energy:
$$
E_{\text {detector }}=E_{\gamma, \text { emitted }} \sqrt{\frac{1+\beta}{1-\beta}}
\tag{46}
$$
where $\beta=v / c$ and $v$ is the velocity of emitter (the ${ }^{58} \mathrm{Ni}$ nucleus). Taking the calculated value of the ${ }^{58} \mathrm{Ni}$ velocity (equation 29 ) we get:
$$
E_{\text {detector }}=E_{\gamma, \text { emitted }} \sqrt{\frac{1+\beta}{1-\beta}}=10.863 \sqrt{\frac{1+0.00565}{1-0.00565}}=10.925 \mathrm{MeV}
\tag{47}
$$

Final answer: 10.8633, 10.925 MeV