The orbital acceleration of an electron in a | vedclass

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Question

(A) According to Bohr's postulate,the angular momentum of an electron is quantized as: $mvr = \frac{nh}{2\pi}$.From this,the orbital velocity $v$ is g

Vedclass · Accepted Answer

$\frac{n^2 h^2}{4 \pi^2 m^2 r^3}$

Vedclass · Answer

(A) According to Bohr's postulate,the angular momentum of an electron is quantized as: $mvr = \frac{nh}{2\pi}$.From this,the orbital velocity $v$ is given by: $v = \frac{nh}{2\pi mr}$.The centripetal (orbital) acceleration $a$ is defined as: $a = \frac{v^2}{r}$.Substituting the expression for $v$ into the acceleration formula:$a = \frac{(\frac{nh}{2\pi mr})^2}{r} = \frac{n^2 h^2}{4 \pi^2 m^2 r^2 \cdot r} = \frac{n^2 h^2}{4 \pi^2 m^2 r^3}$.Thus,the correct option is $A$.

The orbital acceleration of an electron in a hydrogen-like atom is given by:

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