MHT CET · Physics · Atomic Physics

The wavelength of the ﬁrst line in Balmer series in the hydrogen spectrum is ‘ $λ$ ’. What is the wavelength of the second line in the same series?

A $\frac{20}{27} λ$
B $\frac{3}{16} λ$
C $\frac{5}{36} λ$
D $\frac{3}{4} λ$

Verified Solution

Answer & Solution

Correct Answer

(A) $\frac{20}{27} λ$

Step-by-step Solution

Detailed explanation

For Balmer series.

\frac{1}{λ} = R (\frac{1}{2^{2}} - \frac{1}{n^{2}})

\frac{1}{λ} = R (\frac{1}{4} - \frac{1}{9})

\frac{1}{λ} = \frac{5 R}{36}

…(1)
For second line

n = 4

\frac{1}{λ^{1}} = R (\frac{1}{4} - \frac{1}{16}) = \frac{3 R}{16}

…(2)
By equation (1)

\div

(2)

\frac{λ^{1}}{λ} = \frac{5}{38} \times \frac{16}{3}

\Rightarrow λ^{1} = \frac{20}{27} λ

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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The wavelength of the ﬁrst line in Balmer series in the hydrogen spectrum is ‘ λ ’. What is the wavelength of the second line in the same series?

Step-by-step Solution

The wavelength of the ﬁrst line in Balmer series in the hydrogen spectrum is ‘ $λ$ ’. What is the wavelength of the second line in the same series?