The acceleration of electron in the first orbits of hydrogen atom is

Options

A. `h^2/(pi^2 m^2 r^3)`
B. `h^2/(8pi^2 m^2r^3)`
C. `h^2/(4pi^2 m^2 r^3)`
D. `h^2/(4pim^2r^3)`

Answer

C

Explanation

For Hydrogen atom , n=1 and the principal quantum number of first orbit is , n=1
`v=e^2/(2epsilon_0h)`
Radius of the first orbit of the hydrogen atom is given by
`r=(h^2epsilon_0)/(pime^2)`
`rArr epsilon_0=(rpime^2)/h^2`…(i)
The electron revolves in a circular orbit, the acceleration of the electron can be calculated as
`a=v^2/r=e^4/(4epsilon_0^2h^2)xx(pime^2)/(h^2 epsilon_0)=(e^4xxpime^2)/(4h^4 epsilon_0^3)`...(ii)
Substituting `epsilon_0` from equation (i)
Thus , the acceleration of the electron in the first orbit is given by `=(e^4pime^2h^6)/(4h^4r^3pi^3m^3e^6)=h^2/(4pi^2m^2r^3)`