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Q. As the quantum number increases, the difference of energy between consecutive energy levels

Gujarat CETGujarat CET 2006Atoms

Solution:

Total energy in nth Bohr's orbit
$ E_n=- \frac {13.6}{n^2}eV$
where n is the principal quantum number.
$ E_2-E_1=-13.6 \bigg (\frac {1}{(1)^2}- \frac {1}{(2)^2}\bigg )$
$ =-13.6 \bigg (\frac {1}{4}- \frac {1}{1}\bigg )= \frac {13.6 \times 3}{4}$
$ =0.75 \times 13.6 eV$
$ E_3-E_2=-13.6 \bigg (\frac {1}{3^2}- \frac {1}{2^2} \bigg )$
$ =-13.6 \bigg (\frac {1}{9}- \frac {1}{4}\bigg ) $
$ =13.6 \times \frac {5}{36}$
$=0.14 \times 13.6 eV$
Therefore, the energy difference between consecutive energy levels decreases with increase in quantum number.