An electron in hydrogen atom first jumps from second excited state and then from first excited state to ground state. Let the ratio of wavelength, momentum and energy of photons emitted in these two cases be a, b and c respectively. Then

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$c = \frac{1}{a}$

$c = \frac{5}{27}$

$a = \frac{9}{4}$

$b = \frac{5}{27}$

answer is A, C, D.

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Detailed Solution

$E = \frac{h c}{λ} o r E α \frac{1}{λ} and P = \frac{E}{C} o r P α E$

$or \frac{P_{1}}{P_{2}} = \frac{E_{1}}{E_{2}} \Rightarrow b = c$

$\frac{E_{1}}{E_{2}} = \frac{λ_{2}}{λ_{1}} \Rightarrow a = \frac{1}{b} \frac{Δ E_{1}}{Δ E_{2}} = \frac{5}{36} \times \frac{4}{3} = \frac{5}{27} ∴ C = \frac{5}{27} \times b a = \frac{1}{b} = \frac{1}{c}$

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