Question

The ground state energy of hydrogen atom is $-13.6eV$. If an electron makes a transition from an energy level $-0.85eV$ to $-3.4eV$, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does the wavelength belong? ($R=1.097\times {10}^7{m}^{-1}$)

Answer 1

For hydrogen atom we know that $E_n=-\frac{13.6}{n^2}$
Now for the transition from $E_{n_1}=-0.85$ to $E_{n_2}=-3.4$,
$-0.85=-13.6/n_1^2$ or $n_1^2=16$
or $n_1=4$ and $-3.4=-13.6/n_2^2$
or $n_2=2$
Using, $\frac{1}{\lambda }=R(1/n_2^2-1/n_1^2)$
we get $\frac{1}{\lambda }=(1.097\times {10}^7{m}^{-1})(1/2^2-1/4^2)$
or $\lambda =4.86\times {10}^{-7}m$
As the electron make the transition from $n=4$ to $n=2$ so it will be Balmer series.