An electron is moving in Bohr's fourth orbit. Its de-Broglie wave length is λ . What is the circumference of the fourth orbit?

Correct answer: C. 4 λ According to Bohr's concept, an electron always move in the orbit with angular momentum (mvr) equal to n h / 2 π . m v r=(n h)/(2 π) or r=(n)/(2 π) ·((h)/(m v)) Or r=(n λ)/(2 π) (from de-Broglie equation, λ=(h)/(m v) ) For fourth orbit (n=4) , gathered r=(2 λ)/(π) Circumference =2 π r=2 π × (2 λ)/(π)=4 λ gathered

JIPMER 2014

Exam: JIPMER
Year: 2014
Subject: Chemistry
Chapter: Structure of Atom

JIPMER 2014 · Chemistry · Structure of Atom

An electron is moving in Bohr's fourth orbit. Its de-Broglie wave...

An electron is moving in Bohr's fourth orbit. Its de-Broglie wave length is

\lambda

. What is the circumference of the fourth orbit?

Options

A $\frac{2}{\lambda}$
B $2 \lambda$
C $4 \lambda$
D $\frac{4}{\lambda}$

Correct answer

C. $4 \lambda$

Step-by-step solution

According to Bohr's concept, an electron always move in the orbit with angular momentum (mvr) equal to

n h / 2 \pi

\therefore \quad m v r=\frac{n h}{2 \pi}

r=\frac{n}{2 \pi} \cdot\left(\frac{h}{m v}\right) Or r=\frac{n \lambda}{2 \pi}

(from de-Broglie equation,

\lambda=\frac{h}{m v}

) For fourth orbit

(n=4)

\begin{gathered}r=\frac{2 \lambda}{\pi} \\ \therefore \text{ Circumference }=2 \pi r=2 \pi \times \frac{2 \lambda}{\pi}=4 \lambda\end{gathered}

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