What is the wave number of 4^ th line in Balmer series of hydrogen spectrum? (R=1,09,677 cm^ -1 . )

Correct answer: D. 24,372 ~cm^ -1 v = R((1)/(n_1^2)-(1)/(n_2^2)) For Balmer series n_1=2 and for 4^ th line in Balmer series n_2=6 aligned R & =109677 ~cm^ -1 v & =109677((1)/(2^2)-(1)/(6^2)) & =109677((1)/(4)-(1)/(36)) v & =24,372 ~cm^ -1 aligned

JIPMER 2009

Exam: JIPMER
Year: 2009
Subject: Chemistry
Chapter: Structure of Atom

JIPMER 2009 · Chemistry · Structure of Atom

What is the wave number of 4^ th line in Balmer series of hydrogen...

What is the wave number of

4^{\text{th }}

line in Balmer series of hydrogen spectrum?

\left(R=1,09,677 \mathrm{ cm}^{-1}\right. \text{ ) }

Options

A $24,630 \mathrm{~cm}^{-1}$
B $24,360 \mathrm{~cm}^{-1}$
C $24,730 \mathrm{~cm}^{-1}$
D $24,372 \mathrm{~cm}^{-1}$

Correct answer

D. $24,372 \mathrm{~cm}^{-1}$

Step-by-step solution

\bar{v}=

R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)

For Balmer series

n_1=2

and for

4^{\text{th }}

line in Balmer series

n_2=6

\begin{aligned} \because \quad R & =109677 \mathrm{~cm}^{-1} \\ \bar{v} & =109677\left(\frac{1}{2^2}-\frac{1}{6^2}\right) \\ & =109677\left(\frac{1}{4}-\frac{1}{36}\right) \\ \bar{v} & =24,372 \mathrm{~cm}^{-1} \end{aligned}

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