Question

Electrons of energies 10.2 eV, 12.09 eV, and 13.06 eV colliding with a hydrogen atom, can cause radiation to be emitted from the latter. Calculate in each case the principle quantum number, n, 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

The energy of the shell and the principal quantum number are related as
En = - 13.6 eV / n2
The energy needed for the electron to stay at the first shell
E1 = - 13.6 eV / 12 = -13.6 eV
For second shell.
E2 = - 13.6 eV / 22 = -3.4 eV
For third shell,
E3 = - 13.6 eV / 32 = - 1.51 eV
i) for electron energy 10.2 eV
The energy imparted by the electron to the Hydrogen atom is used to make transition from n = 1 to n = 2
E2 - E1 = -3.4 eV - (-13.6 eV) =10.2 eV
Thus the hydrogen atom moves from the ground state to n = 1 shell.
En = 10.2 eV = 13.6 eV / n2
ii)) For the electron energy 12.09 eV
The Hydrogen atom uses this energy to make transition from n = 1 to n = 3
E3 - E1 = - 1.51 eV - (-13.6 eV ) = 12.09 eV
iii) For the electron energy 13.6 eV
This is the first ionization energy of the hydrogen atom. So the atom loses its electron.
b) The wavelength of the radiation emitted
i) 10.2 eV
E = h c /
= h c / E
= 6.626 x 10-34 Js x 3 x 108 m/s / 10.2 x 1.6 x 10-19 J
=  1.2180 x 10-7 m = 121.8 nm
ii) 12.09 eV
= h c / E
= 6.626 x 10-34 Js x 3 x 108 m/s / 12.09 x 1.6 x 10-19 J
=  1.0276 x 10-7 m = 102.76 nm
iii) 13.6 eV
= h c / E
= 6.626 x 10-34 Js x 3 x 108 m/s / 13.6 x 1.6 x 10-19 J
=  0.9135 x 10-7 m = 91.35 nm