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When an electron jumps from 2^nd to 1^st , 3^rd to 2^nd and 3^rd to 1^st shells, the wave lengths of radiations emitted are $λ_{1}, λ_{2}$ and $λ_{3}$ respectively. The relationship between $λ_{1}; λ_{2}$ and $λ_{3}$ is

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$λ_{3} = \frac{λ_{1} + λ_{2}}{λ_{1} . λ_{2}}$

$λ_{3} = λ_{1} + λ_{2}$

$λ_{3} = \frac{λ_{1} . λ_{2}}{λ_{1} + λ_{2}}$

$λ_{3} = {[λ_{1} + λ_{2}]}^{2}$

answer is C.

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Detailed Solution

$\frac{h c}{λ_{3}} = \frac{h c}{λ_{2}} + \frac{h c}{λ_{1}} \frac{1}{λ_{3}} = \frac{1}{λ_{2}} + \frac{1}{λ_{1}} = \frac{λ_{2} + λ_{1}}{λ_{1} λ_{2}} λ_{3} = \frac{λ_{1} . λ_{2}}{λ_{1} + λ_{2}}$

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