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(D) The magnetic force provides the centripetal force for the circular motion: $F_m = F_c \Rightarrow evB = \frac{mv^2}{r} \Rightarrow mv = eBr$. Acco

Vedclass · Accepted Answer

$\sqrt{\frac{h}{\pi e B}}$

Vedclass · Answer

(D) The magnetic force provides the centripetal force for the circular motion: $F_m = F_c \Rightarrow evB = \frac{mv^2}{r} \Rightarrow mv = eBr$. According to Bohr's quantization condition for angular momentum: $mvr = \frac{nh}{2\pi}$. Substituting $mv = eBr$ into the quantization condition: $(eBr)r = \frac{nh}{2\pi} \Rightarrow er^2B = \frac{nh}{2\pi}$. Solving for $r$: $r = \sqrt{\frac{nh}{2\pi eB}}$. For the first excited state,the principal quantum number is $n = 2$. Substituting $n = 2$ into the expression: $r = \sqrt{\frac{2h}{2\pi eB}} = \sqrt{\frac{h}{\pi eB}}$.

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 $:$

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