Answer
Beryllium.
Work Step by Step
Use equation 27–16 for the wavelength.
$$\frac{1}{\lambda}=(constant)Z^2\left( \frac{1}{n_x’^{2}}-\frac{1}{n_x^{2}}\right)$$
In this problem, we are looking for an element X, with atomic number Z. The condition is that a final state $n_X’$, for every fourth value of initial state $n_X$, the wavelength equals a wavelength in the Lyman series(for some initial state in hydrogen $n_H$).
Equate the wavelengths in the spectrum for hydrogen and for the unknown element.
$$\left( \frac{1}{1^{2}}-\frac{1}{n_H^{2}}\right)= Z^2\left( \frac{1}{n_x’^{2}}-\frac{1}{n_x^{2}}\right)$$
$$\left( \frac{1}{1^{2}}-\frac{1}{n_H^{2}}\right)= \frac{Z^2}{n_x’^{2}}-\frac{Z^2}{n_x^{2}}$$
Consider the first term on each side. We obtain $n_x’=Z$ for the final n.
Now consider the second term, which involves the initial n.
$$\frac{1}{n_H^{2}}= \frac{Z^2}{n_x^{2}}$$
$$Z^2n_H^2=n_x^2$$
$$Zn_H= n_x$$
This relation holds true only for every fourth line, i.e., every fourth value of $n_X$. This condition is met for Z = 4, in which case the first condition tells us that $n_x’=4$.
Since Z=4, the element is beryllium.
In summary, the spectral lines for $n_{Be}=4, 8, 12,…$ will match the hydrogen Lyman series lines for $n_H=1, 2, 3,…$.
