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AIIMS 2014

Exam: AIIMS
Year: 2014
Subject: Physics
Chapter: Atomic Physics

AIIMS 2014 · Physics · Atomic Physics

If the wavelength of 1^ st line of Balmer series of hydrogen is...

If the wavelength of

1^{\text{st }}

line of Balmer series of hydrogen is

6561 Å

, the wavelength of the

2^{\text{nd }}

line of series will be

Options

A $9780 Å$
B $4860 Å$
C $8857 Å$
D $4429 Å$

Correct answer

B. $4860 Å$

Step-by-step solution

For the first line of Balmer series,

\frac{1}{\lambda_1}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{5 R}{36}

For the second line of Balmer series.

\frac{1}{\lambda_2}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{3 R}{16}

\therefore \frac{\lambda_2}{\lambda_1}=\frac{5 R / 36}{3 R / 16}=\frac{20}{27}

; or

\lambda_2=\frac{20}{27}(6561 Å)=4860 Å

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