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JEE Main 201212 MayOnline

Exam: JEE Main
Year: 2012
Date: 12 May
Mode: Online
Subject: Physics
Chapter: Atomic Physics

JEE Main 2012 · 12 May · Online · Physics · Atomic Physics

A doubly ionised Li atom is excited from its ground state(n=1) to...

A doubly ionised

\mathrm{Li}

atom is excited from its ground

\operatorname{state}(n=1)

to

n=3

state. The wavelengths of the spectral lines are given by

\lambda_{32}, \lambda_{31}

and

\lambda_{21}

. The ratio

\lambda_{32} / \lambda_{31}

and

\lambda_{21} / \lambda_{31}

are, respectively

Options

A $8.1,0.67$
B $8.1,1.2$
C $6.4,1.2$
D $6.4,0.67$

Correct answer

C. $6.4,1.2$

Step-by-step solution

\frac{1}{\lambda}=R

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

where

R=

Rydberg

$\begin{aligned} & \text{ constant } \\ & \frac{1}{\lambda_{32}}=\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36} \\ & \Rightarrow \lambda_{32}=\frac{36}{5} \end{aligned}

Similarly solving for

\lambda_{31}

and

\lambda_{21}

\begin{aligned} & \lambda_{31}=\frac{9}{8} \text{ and } \lambda_{21}=\frac{4}{3}\\ & \therefore \quad \frac{\lambda_{32}}{\lambda_{31}}=6.4 \text{ and } \frac{\lambda_{21}}{\lambda_{31}} \simeq1.2 \end{aligned}

$

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