The ionization energy of the electron in the hydrogen atom in its ground state is 13.6 eV. The atoms are excited to higher energy levels to emit radiations of 6 wavelengths. Maximum wavelength of emitted radiation corresponds to the transition between
- n = 3 to n = 1 states
- n = 2 to n = 1 states
- n = 4 to n = 3 states
- n = 3 to n = 2 states
Answer
(3) n = 4 to n = 3 states
Solution
Number of spectral lines from level n: n(n-1)2 = 6
n = 4
Maximum wavelength minimum energy difference, i.e. between the two highest levels:
E4 3 = 13.6 ( 132 - 142 ) = 0.66 eV (smallest)
The transition is n = 4 to n = 3.
Answer: (3)