The energy of an electron in the nth Bohr orbit of hydrogen atom is

Explanation

Consider the electron revolving in the nth orbit around the hydrogen nucleus. Let m and -e be the mass and the charge of the electron, r the radius of th eorbit and v, the liner speed of the `v`, the linear speed of the electron.
According to Bohr's first postulate, centripetal force acting on the electron = electrostatic force of attraction exerted on the electron by the nucleus
`therefore (mv^(2))/(r)=(1)/(4pi epsilon_(0))*(e^(2))/(r^(2)) " " ` ... (1)
where `epsilon_(0)` is the permittivity of free space.
`therefore` Kinetic energy (KE) of the electron `=(1)/(2) mv^(2)=(e^(2))/(8pi epsilon r) " " ` ...(2)
The electric potential due to the hydrogen nucleus (charge = +e) at a point at a distance r from it is `V=(1)/(4piepsilon_(0))*(e)/(r)`
`therefore` Potential energy (PE) of the electron
=charge on the electron `xx` electric potential
`=-e xx (1)/(4pi epsilon_(0))(e)/(r)= -(e^(2))/(4piepsilon_(0)r) " " ` ...(3)
Hence, the total energy of the electron in the nth orbit is
`E=KE+PE =(-e^(2))/(4pi epsilon_(0)r)+(e^(2))/(8pi epsilon_(0)r)`
`therefore E= -(e^(2))/(8pi epsilon_(0)r) " "` ...(4)
This shows that the total energy of the electron in the nth orbit of hydrogen atom is inversely proportional to the radius of the orbit as `epsilon_(0)` and e are constants.
The radius of the nth orbit of the electron is
`r=(epsilon_(0)h^(2)n^(2))/(pi me^(2)) " " ` ...(5)
where h is Planck's constant.
From Eqs. (4) and (5), the energy of the electron in the nth Bohr orbit is
`E_(n)=-(e^(2))/(8pi epsilon_(0))((pi me^(2))/(epsilon_(0)h^(2)n^(2)))`
`= (me^(4))/(8epsilon_(0)^(2)h^(2)n^(2)) " "`...(6)
The minus sign in the shows that the electron is bound to the nucleus by the electrostatic force of attraction.
As m, e, `epsilon_(0)` and h are constant, we get,
`E_(n) prop (1)/(n^(2))`
i.e., the energy of the electron in a stationary energy state is discrete and is inversely proportional to the square of the principal quantum number.