MHT CET · Physics · Atomic Physics

In Balmer series, wavelength of first line is $' λ_{1}^{'}$ and in Brackett series wavelength of first line is $' λ_{2}^{'}$ then $\frac{λ_{1}}{λ_{2}}$ is

A 0.162
B 0.124
C 0.138
D 0.188

Verified Solution

Answer & Solution

Correct Answer

(A) 0.162

Step-by-step Solution

Detailed explanation

The wavelength of a line in Balmer series is given by

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

[for n = 3, 4, 5,….]
Where, R= Rydberg constant.
For first line, n = 3

\Rightarrow \frac{1}{λ_{1}} = R (\frac{1}{4} - \frac{1}{9}) \Rightarrow λ_{1} = \frac{35}{5 R}

….(i)
The wavelength of a line in Brackett series is given by

\frac{1}{λ} = R (\frac{1}{4^{2}} - \frac{1}{n^{2}})

[for n = 5, 6, 7]
For first line, n = 5

\Rightarrow \frac{1}{λ_{2}} = R (\frac{1}{16} - \frac{1}{25}) \Rightarrow λ_{2} = \frac{400}{9 R}

……(ii)
Dividing Eq. (i) by Eq. (ii), we get

\frac{λ_{1}}{λ_{2}} = \frac{36}{5 R} \times \frac{9 R}{400} = \frac{81}{500} = 0.162

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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In Balmer series, wavelength of first line is 'λ1' and in Brackett series wavelength of first line is 'λ2' then λ1λ2 is

Step-by-step Solution

In Balmer series, wavelength of first line is $' λ_{1}^{'}$ and in Brackett series wavelength of first line is $' λ_{2}^{'}$ then $\frac{λ_{1}}{λ_{2}}$ is