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The shortest wave length of H-atom in the Lyman series is $λ 1$ , The longest wavelength in the Balmer series of $He +$ is.

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36 λ 1 5

5 λ 1 9

9 λ 1 5

27 λ 1 5

answer is C.

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Detailed Solution

Shortest wavelength

→ Max .energy (x → 1)

For Lyman series of Hatom.

1 λ 1 = R H (1) 2 11 − 0 ⇒ 1 λ 1 = R H ⇒ R H = 1 λ 1

For Balmer series of

He + .

1 λ = R H (2) 2 122 − 132 ⇒ 1 λ = R H (4) 9 − 4 36 ⇒ 1 λ = 5 R H 9 ⇒ λ = 9 5 R H = 9 λ 1 5

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