Atomic StructureHard

Question

The angular momentum of electron in Bohr’s orbit is J. What will be the K.E. of electron in that Bohr’s orbit? (a) $\frac{1}{2}\frac{Jv}{r}$ (b) $\frac{Jv}{r}$ (c) $\frac{J^{2}}{2m}$ (d) $\frac{J^{2}}{2r}$

Options

A.$\frac{1}{2}\frac{Jv}{r}$
B.$\frac{Jv}{r}$
C.$\frac{J^{2}}{2m}$
D.$\frac{J^{2}}{2r}$

Solution

$K.E. = \frac{1}{2}mv^{2} = \frac{1}{2}.\frac{(mvr)v}{r} = \frac{J.V}{2r}$

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