Question

The ground state energy of hydrogen atom is $-13.6$ eV. If an electron makes a transition from an energy level $-1.51$ eV to $-3.4$ eV, then calculate the wavelength of the spectral line emitted and name the series of hydrogen spectrum to which it belongs.

Answer 1

Given ground state energy of hydrogen $E_n$=$\frac{-13.6}{n^2}eV$

As per question there is only 1 hydrogen atom than $E_1$=$-13.6$ eV

Now during electron transition from from one energy level to another energy level

$E_2=-1.51eVand{E}_3=-3.4eV$

So, $E_1=E_2$

$-1.51=\frac{-13.6}{{n^2}_1}$

$n_2=3$,similarly solving for $E_2$ we get the value of $n_3$=$2$

Thus electron transition takes place from $n=3$ to $n=2$ which corresponds to Balmer series

Thus $\frac{1}{\lambda }=R\{\frac{1}{n_2^3}-\frac{1}{n_2^2}\}$

Substituting the values in above equation we get

$\frac{1}{\lambda }=1.097\times {10}^7\{\frac{1}{4}-\frac{1}{9}\}$

$\lambda =6563\mathring{A}$