Question 272: Atoms & Nuclei

Question

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 :

Accepted Answer

n = 1 (Ground state)
n = 2 (First excitate state)
n = 3 (Second excitate state)
n = 4 (Third excitate state)
$${{hc} \over {{\lambda _1}}} = 13.6\left( {{1 \over 9} - {1 \over {16}}} ight)$$
$${{hc} \over {{\lambda _2}}} = 13.6\left( {{1 \over 4} - {1 \over 9}} ight)$$
$${{{\lambda _2}} \over {{\lambda _1}}} = {{\left( {{7 \over {9 	imes 16}}} ight)} \over {\left( {{5 \over {9 	imes 4}}} ight)}}$$ = $${7 \over {20}}$$
$$ 	herefore $$ $${{{\lambda _1}} \over {{\lambda _2}}} = {{20} \over 7}$$

Answer

$${{22} \over 5}$$

Answer

$${7 \over 5}$$

Answer

$${9 \over 7}$$

Answer

$${{20} \over 7}$$

Atoms & Nuclei

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 :

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