Question

What transition in the hydrogen spectrum would have the same wavelength as the Balmer transition n = 4 to n = 2 of He⁺ spectrum?

Answer 1

Concept Used:

For hydrogen-like species, the wavelength of a spectral line is given by:

1/λ = R Z² ( 1/n₁² − 1/n₂² )

where Z is the atomic number.

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Step 1: Write expression for He⁺ transition

For He⁺, Z = 2

Given transition: n = 4 → n = 2

1/λ = R (2)² ( 1/2² − 1/4² )

1/λ = 4R ( 1/4 − 1/16 )

1/λ = 4R ( 3/16 )

1/λ = 3R/4

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Step 2: Find equivalent hydrogen transition

For hydrogen, Z = 1

1/λ = R ( 1/n₁² − 1/n₂² )

Equating with 3R/4:

1/n₁² − 1/n₂² = 3/4

This condition is satisfied for:

n₁ = 1 and n₂ = 2

since,

1 − 1/4 = 3/4

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Final Answer:

The hydrogen transition having the same wavelength is:

n = 2 → n = 1 (Lyman series)