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JEE Main 20127 MayOnline

Exam: JEE Main
Year: 2012
Date: 7 May
Mode: Online
Subject: Physics
Chapter: Atomic Physics

JEE Main 2012 · 7 May · Online · Physics · Atomic Physics

The electron of a hydrogen atom makes a transition from the (n+1)^...

The electron of a hydrogen atom makes a transition from the

(n+1)^{\text{th }}

orbit to the

n^{\text{th }}

orbit. For large

\mathrm{n}

the wavelength of the emitted radiation is proportional to

Options

A $n$
B $n^3$
C $n^4$
D $n^2$

Correct answer

B. $n^3$

Step-by-step solution

If

n_1=n

and

n_2=n+1

Maximum wavelength

\lambda_{\max }=\frac{n^2\left(n+1^2\right)}{(2 n+1 R)}

Therefore, for large

\mathrm{n}, \lambda_{\max } \propto n^3

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