Question
Which of the following transitions in hydrogen atoms emit photons of highest frequency?
A.
$$n = 1\,{\text{to}}\,n = 2$$
B.
$$n = 2\,{\text{to}}\,n = 6$$
C.
$$n = 6\,{\text{to}}\,n = 2$$
D.
$$n = 2\,{\text{to}}\,n = 1$$
Answer :
$$n = 2\,{\text{to}}\,n = 1$$
Solution :
Frequency is given by
$$hv = - 13.6\left( {\frac{1}{{n_2^2}} - \frac{1}{{n_1^2}}} \right)$$
For transition from $$n = 6\,{\text{to}}\,n = 2,$$
$${v_1} = \frac{{ - 13.6}}{h}\left( {\frac{1}{{{6^2}}} - \frac{1}{{{2^2}}}} \right) = \frac{2}{9} \times \left( {\frac{{13.6}}{h}} \right)$$
For transition from $$n = 2\,{\text{to}}\,n = 1,$$
$$\eqalign{
& {v_2} = \frac{{ - 13.6}}{h}\left( {\frac{1}{{{2^2}}} - \frac{1}{{{1^2}}}} \right) = \frac{3}{4} \times \left( {\frac{{13.6}}{h}} \right) \cr
& \therefore {v_1} > {v_2} \cr} $$