JEE MAIN - Physics (2007 - No. 3)
Which of the following transitions in hydrogen atoms emit photons of highest frequency ?
$$n = 1$$ to $$n=2$$
$$n = 2$$ to $$n=6$$
$$n = 6$$ to $$n=2$$
$$n = 2$$ to $$n=1$$
Explanation
We have no find the frequency of emitted photons. For emission of photons the transition must take place from a higher energy level to a lower energy level which are given only in options $$(c)$$ and $$(d)$$.
Frequency is given by
$$hv = - 13.6\left( {{1 \over {n_2^2}} - {1 \over {n_1^2}}} \right)$$
For transition from $$n=6$$ to $$n=2,$$
$${v_1} = {{ - 13.6} \over h}\left( {{1 \over {{6^2}}} - {1 \over {{2^2}}}} \right)$$
$$ = {2 \over 9} \times \left( {{{13.6} \over h}} \right)$$
For transition from $$n=2$$ to $$n=1,$$
$${v_2} = {{ - 13.6} \over h}\left( {{1 \over {{2^2}}} - {1 \over {{1^2}}}} \right)$$
$$ = {3 \over 4} \times \left( {{{13.6} \over h}} \right).$$
$$\therefore$$ $${v_1} > {v_2}$$
Frequency is given by
$$hv = - 13.6\left( {{1 \over {n_2^2}} - {1 \over {n_1^2}}} \right)$$
For transition from $$n=6$$ to $$n=2,$$
$${v_1} = {{ - 13.6} \over h}\left( {{1 \over {{6^2}}} - {1 \over {{2^2}}}} \right)$$
$$ = {2 \over 9} \times \left( {{{13.6} \over h}} \right)$$
For transition from $$n=2$$ to $$n=1,$$
$${v_2} = {{ - 13.6} \over h}\left( {{1 \over {{2^2}}} - {1 \over {{1^2}}}} \right)$$
$$ = {3 \over 4} \times \left( {{{13.6} \over h}} \right).$$
$$\therefore$$ $${v_1} > {v_2}$$
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