Question

Electrons of energies $10.2eV$ and $12.09eV$can cause radiation to be emitted from hydrogen atom. Calculate in each case the principal quantum number of the orbit to which the electron in the hydrogen atom is raised and the wavelength of the radiation emitted if it drops back to the ground state.

Answer 1

$E_n^H=\frac{13.6}{n^2}eV$
$\therefore E_1^H=\frac{13.6}{1^2}eV=-13.6eV(\because n=1)$
$E_2^H=\frac{13.6}{2^2}eV=-13.4eV(\because n=2)$
$E_3^H=\frac{13.6}{3^2}eV=-1.51eV(\because n=3)$
Now, $E_2^H-E_1^H=-3.4eV+13.6eV=10.2eV$
$E_3^H-E_1^H=-1.51eV+13.6eV=12.09eV$

Hence electron will be raised to the state of principal quantum number $2$ and $3$ corresponding to radiation energies $10.2eV$ and $12.09eV$ respectively. Wavelength of radiation emitted for the first is

$\lambda =\frac{c}{v}=\frac{hc}{\Delta E}=\frac{6.62\times {10}^{-34}\times 3\times {10}^6}{10.2\times 1.6\times {10}^{-19}}=1217\times {10}^{-10}meter=1217\mathring{A}$

Similarly, for the other $\lambda =\frac{6.62\times {10}^{-34}\times 3\times {10}^6}{10.2\times 1.6\times {10}^{-19}}=1026\mathring{A}.$