Question

How many hydrogen atoms in the ground state are excited by means of monochromatic radiation of wavelength $970.6\mathring{A}$. How many different lines are possible in the resulting emission spectrum? Find the longest wavelength among these.

Answer 1

Six different lines are possible,

$\lambda =1215.68\mathring{A}$.
$E_n-E_1=\frac{hc}{\lambda }$
$\frac{-21.69\times {10}^{-19}}{n^2}+\frac{21.69\times {10}^{-19}}{1}=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{970.6\times {10}^{-10}}$
$n=4$
$\frac{1}{\lambda }=R{Z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
$\frac{1}{\lambda }=109677.77\times 1^2(\frac{1}{1^2}-\frac{1}{2^2})$
$\lambda =1215.68\mathring{A}$. is the longest wavelength among all six.