Question

The wavelength of the first line of Lyman series for hydrogen is identical to that of the second line of Balmer series for some hydrogen-like ion $X$. Calculate energies of the first four levels of $X$. Also, find its ionization potential. (Given: Ground state binding energy of the hydrogen atom is $13.6eV$)

Answer 1

Wavelength of first line of Lyman series : $\frac{1}{{\lambda }_1}=R{Z}_1^2(1/1^2-1/2^2)=3R/4$ (for hydrogen Z=1)

Wavelength of second line of Balmer series of element X : $\frac{1}{{\lambda }_2}=R{Z}_2^2(1/2^2-1/4^2)=\left(3R/16\right)Z_2^2$

Here, ${\lambda }_1={\lambda }_2$

$\Rightarrow Z_2=2$

Thus, energy levels of X $\left(H{e}^{+}\right)$ are $E_n=-\frac{13.6Z^2}{n^2}eV$

So, $E_1=-13.6\left(4\right)/1^2=-54.4eV,E_2=-13.6\left(4\right)/2^2=-13.6eV,$$E_3=-13.6\left(4\right)/3^2=-6.04eV,E_4=-13.6\left(4\right)/16=-3.52eV$