Question

In a hydrogen atom, the energy of an electron in an orbit is $-3.4$ eV. When it is excited to the next higher energy level and comes back to ground state, calculate the wavelength of emitted radiation.

Answer 1

As we know,
$E=-\frac{13.6}{n^2}\times Z^2,Z=1$ for ${}^{\prime }{H}^{\prime }$ atom
$E=-\frac{13.6}{n_1^2}=-3.4eV$
$n_1=2$
Electron is excited to next level of $n_{1}i.e.,n_2=3$.
Electron transition $n_2=3\to n_3=1$
$n_2=3,E_2=-\frac{21.72\times {10}^{-19}}{9}Joules$

$E_2=-\frac{21.72\times {10}^{-19}}{1}Joules$

$\Delta E=2.414\times {10}^{-19}Joules$

$E=\frac{hc}{\lambda }=\frac{6.314\times {10}^{-34}\times 3\times {10}^8}{\lambda }$

$=2.414\times {10}^{-19}J$
$\lambda =\frac{6.314\times {10}^{-34}\times 3\times {10}^8}{2.414\times {10}^{-19}}$

$\lambda =9\times {10}^{-7}m$