JEE MAIN - Physics (2019 - 12th April Evening Slot - No. 1)
The electron in a hydrogen atom first jumps from the third excited state to the second excited state and
subsequently to the first excited state. The ratio of the respective wavelengths, $${{{\lambda _1}} \over {{\lambda _2}}}$$, of the photons emitted
in this process is :
$${{22} \over 5}$$
$${7 \over 5}$$
$${9 \over 7}$$
$${{20} \over 7}$$
Explanation
n = 1 (Ground state)
n = 2 (First excitate state)
n = 3 (Second excitate state)
n = 4 (Third excitate state)
$${{hc} \over {{\lambda _1}}} = 13.6\left( {{1 \over 9} - {1 \over {16}}} \right)$$
$${{hc} \over {{\lambda _2}}} = 13.6\left( {{1 \over 4} - {1 \over 9}} \right)$$
$${{{\lambda _2}} \over {{\lambda _1}}} = {{\left( {{7 \over {9 \times 16}}} \right)} \over {\left( {{5 \over {9 \times 4}}} \right)}}$$ = $${7 \over {20}}$$
$$ \therefore $$ $${{{\lambda _1}} \over {{\lambda _2}}} = {{20} \over 7}$$
n = 2 (First excitate state)
n = 3 (Second excitate state)
n = 4 (Third excitate state)
$${{hc} \over {{\lambda _1}}} = 13.6\left( {{1 \over 9} - {1 \over {16}}} \right)$$
$${{hc} \over {{\lambda _2}}} = 13.6\left( {{1 \over 4} - {1 \over 9}} \right)$$
$${{{\lambda _2}} \over {{\lambda _1}}} = {{\left( {{7 \over {9 \times 16}}} \right)} \over {\left( {{5 \over {9 \times 4}}} \right)}}$$ = $${7 \over {20}}$$
$$ \therefore $$ $${{{\lambda _1}} \over {{\lambda _2}}} = {{20} \over 7}$$
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