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Q.

In hydrogen atom H_α – line arises due to transition n = 3→ n = 2. In the spectrum of singly ionised helium there is a line having the same wavelength as the H_a line. This is due to the transition

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a

n = 3 → n = 2

b

n = 2 → n = 1

c

n = 5 → n = 3

d

n = 6 → n = 4

answer is D.

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Detailed Solution

$given, λ_{3 \to 2} (H) = λ_{n_{2} \to n_{1}} (He) i . e ., {\bar{ν}}_{3 \to 2} (H) = {\bar{ν}}_{n_{2} \to n_{1}} (He) R (\frac{1}{2^{2}} - \frac{1}{3^{2}}) = R \times 4 (\frac{1}{n_{2}^{2}} - \frac{1}{n_{1}^{2}}) From options, n_{1} = 6 and n_{2} = 4$

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In hydrogen atom Hα – line arises due to transition n = 3→ n = 2. In the spectrum of singly ionised helium there is a line having the same wavelength as the Ha line. This is due to the transition