The kinetic energy of the electron in an orbit of radius `r` in hydrogen atom is (`e =` electronic charge)

Options

A. `(e^(2))/r`
B. `(e^(2))/(2r)`
C. `(e^(2))/r`
D. `(e^(2))/(2r^(2))`

Answer

B

Explanation

Refer fig. shown here ltBrgt
`(mv_(n)^(2))/(r_(n))=1/(4piepsilon_(0)) (Zexxe)/(r_(n)^(2))=(kZe^(2))/(r_(n)^(2))`
`mv_(n)^(2)=(kZe^(2))/(r_(n))`
Kinetic energy of electron in nth orbit is
`K=1/2 mv_(n)^(2)= (kZe^(2))/(2r_(n))`
for hydrogen atom, Z=1 and in cgs system k=1
`K=(e^(2))/(2r_(n))=(e^(2))/(2r)`