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Question

The wavelength of the emitted radiation, if electron in hydrogen atom jumps from the third orbit to second orbit is


A
λ=365R
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B
λ=5R36
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C
λ=5R
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D
λ=R6
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Solution

The correct option is A $$\lambda =\dfrac{36}{5R}$$
Wave number, $$\bar{v}=R\left [ \dfrac{1}{n_{1}^{2}}-\dfrac{1}{n_{2}^{2}} \right ]$$
$$\Rightarrow \lambda =\dfrac{1}{R[\dfrac{1}{4}-\dfrac{1}{9}]}$$
$$\Rightarrow \lambda =\dfrac{36}{5R}$$

Physics

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