Question
Which of the following transitions gives photon of maximum energy?
A.
$$n = 1$$ to $$n = 2$$
B.
$$n = 2$$ to $$n = 1$$
C.
$$n = 2$$ to $$n = 6$$
D.
$$n = 6$$ to $$n = 2$$
Answer :
$$n = 2$$ to $$n = 1$$
Solution :
Energy levels of $$H$$-atom are given by
$$\eqalign{
& {E_n} = - \frac{{13.6{Z^2}}}{{{n^2}}}eV\,\,\left( {Z = 1} \right) \cr
& \Rightarrow {E_n} = \frac{{ - 13.6}}{{{n^2}}}\,eV \cr} $$
Photons are emitted only when electron jumps from higher energy level (higher $$n$$-value) to lower energy level (lower $$n$$-value). So, alternative (a) and (c) are wrong.
Energy difference from $$n = 2$$ to $$n = 1$$ level is
$$\Delta {E_{2 \to 1}} = 13.6\left( {\frac{1}{{{1^2}}} - \frac{1}{{{2^2}}}} \right)eV = 13.6 \times \frac{3}{4} = 10.2\,eV$$
$$\eqalign{
& \Delta {E_{6 \to 2}} = 13.6\left( {\frac{1}{{{2^2}}} - \frac{1}{{{6^2}}}} \right) \cr
& = 13.6 \times \left( {\frac{1}{4} - \frac{1}{{36}}} \right) = 13.6 \times \frac{2}{9} = 3.02\,eV \cr} $$
Thus, it is evident that difference is larger for $$n = 2$$ to $$n = 1$$ transition. Hence, maximum energy photon or shortest wavelength will be emitted during transition from $$n =2$$ to $$n = 1$$