In Bohr's orbit, kinetic energy of an electron in the `n^(th)` orbit of an atom in terms of angular momentum is

Options

A. `1//L`
B. `1//L^(2)`
C. `L^(2)`
D. `1//L^(3)`

Answer

C

Explanation

`K.E.=1/2 mv^(2)`
`=1/2 mv^(2) xxm/m`
`=1/2 (m^(2)v^(2))/m`
`=p^(2)/(2m)" "( :' p=mv)`
`=L^(2)/(2m r^(2))" "( :' P=L/r)`