What is the orbital acceleration of the electron in the first bohr orbit of hydrogen atom ?

Options

A. `(h)/(2pi^2m^2r^3)`
B. `(h^2)/(2pi^2m^2r^2)`
C. `(h^2)/(4pi^2m^2r^3)`
D. `(h^3)/(4pi^2m^2 r^2)`

Answer

C

Explanation

According to Bohr's quantum condition, angular momentum , `mvr = (nh)/(2pi) " and for n=1" , mvr = (h)/(2pi)`
`therefore v= (h)/(2pimr)`
and the orbital acceleration `(a) = v^2/r`
`therefore a= (h^2)/(4pi^2m^2r^2xxr) = (h^2)/(4pi^2m^2r^3)`