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The shortest wavelength of H – atom in Lyman series is x, longest wavelength In Balmer series of He+ is

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9x/5

36x/5

x/4

5x/9

answer is A.

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Detailed Solution

Shortest wavelength means maximum wave number where n₂ = ∞
For the shortest wavelength of H-atom in Lyman series,n₁ = 1 and n₂ = ∞

$large bar v = frac{1}{lambda ^1} = {R_H}left( {frac{1}{{{1^2}}} - frac{1}{{{2^2}}}} right) = {R_H}$

$large {lambda _1} = frac{1}{{{R_H}}} = x......(1)$

Longest wavelength means minimum wave number, Where n₂ = n₁ +1

Thus in Balmar series of He⁺ n₁ = 2 and n₂ = 2 + 1 = 3; Z = 2

$large bar v = frac{1}{lambda^2 } = {R_H} {2^2}left( {frac{1}{{{2^2}}} - frac{1}{{{3^2}}}} right) = 4{R_H}left( {frac{5}{{36}}} right) = frac{{5{R_H}}}{9}$

$large {lambda _2} = frac{9}{{5{R_H}}}......(2)$

Substituting (1) in (2)

$large {lambda _2} = frac{9}{5}left( {frac{1}{{{R_H}}}} right);{lambda _2} = frac{{9x}}{5}$

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