A hydrogen-like atom with atomic number $Z$ is in an excited state of principal quantum number $n$. It emits a photon of wavelength $\lambda$ when it transitions to the ground state. If the recoil momentum of the atom is $p$, then the value of $Z$ is given by which expression?

Solution

1. The momentum of the emitted photon is given by $p=\frac{h}{\lambda }$, and the energy of the photon is $$\begin{array}{rl}E& =pc\\ & =\frac{hc}{\lambda }\end{array}$$.
2. According to the Rydberg formula for a hydrogen-like atom, the energy of the emitted photon is $E=13.6Z^2\left(1-\frac{1}{n^2}\right)\text{ eV}$, which can be expressed as $E=R_{H}hc{Z}^2\left(\frac{n^2-1}{n^2}\right)$.
3. Equating the two expressions for energy, $\frac{hc}{\lambda }=R_{H}hc{Z}^2\left(\frac{n^2-1}{n^2}\right)$, and substituting $R_H=\frac{1}{{\lambda }_0}$ where ${\lambda }_0$ relates to the momentum, we solve for $Z$.
4. Rearranging the equation $\frac{1}{\lambda }=R_H{Z}^2\left(\frac{n^2-1}{n^2}\right)$ and incorporating the recoil momentum $p=\frac{h}{\lambda }$ leads to $Z=\frac{p\lambda }{h}\left(\frac{n^2}{n^2-1}\right)$.
5. Hence the answer is (A).