An electron projected perpendicular to a uniform magnetic field B… - Vidyadip

MCQ

An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr's quantization is applicable, then the radius of the electronic orbit in the first excited state is :

A
$\sqrt{\frac{2 h}{\pi e B}}$
B
$\sqrt{\frac{4 h}{\pi eB }}$
C
$\sqrt{\frac{ h }{2 \pi eB }}$
✓
$\sqrt{\frac{ h }{\pi eB }}$

✓

Answer

Correct option: D.

$\sqrt{\frac{ h }{\pi eB }}$

(D) $\sqrt{\frac{ h }{\pi eB }}$
Sol. $r =\frac{ mv }{ eB } \& mvr =\frac{ nh }{2 \pi} \Rightarrow( eBr ) r =\frac{ nh }{2 \pi}$
$\Rightarrow r =\sqrt{\frac{ nh }{2 \pi eB }}$
first excited state $: n=2 \therefore r=\sqrt{\frac{h}{\pi e B}}$

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