Question

Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength $975\stackrel{\circ }{A}$. How many different lines are possible in the resulting spectrum? Calculate the longest wavelength amongst them. You may assume the ionization energy for hydrogen atom as $13.6eV$.
(Planck's constant $=6.63\times {10}^{-34}J-s,c=3\times {10}^{8}m/s$).

Answer 1

Given that,

$\lambda =975\dot{A}$

Ionization energy of hydrogen atom $=13.6eV$

$E=\frac{hc}{\lambda }$

$E=\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{975\times {10}^{-10}\times 1.6\times {10}^{-19}}$

$E=12.75eV$

$\therefore 12.75=13.6(\frac{1}{1^2}-\frac{1}{n^2})$

$n=4$

So, the number of lines in the resultant spectrum is 6. The longest wavelength is emitted from 4^th orbit to 3^rd orbit. i.e.

$E_{4\to 3}=13.6(\frac{1}{3^2}-\frac{1}{4^2})$

Longest wavelength,

$\lambda =\frac{hc}{E_{4\to 3}}$

$\lambda =\frac{6.6\times {10}^{-34}\times 3\times {10}^8}{13.6\times 7\times 1.6\times {10}^{-19}}$

$\lambda =1.88\times {10}^{-6}m$

$\lambda =18800\dot{A}$