Derive the energy expression for hydrogen atom using Bohr atom model.

Explanation

Since the electrostatic force is a conservative force , the potential energy for the `n^(th)` orbit is
`U_n= 1/(4pivarepsilon_0)((+Ze)(-e))/(r_n)=-1/(4pivarepsilon_0)(Ze^2)/(r_n)`
`=-1/(4varepsilon_0^2)(Z^2me^4)/(h^2n^2)(because r_n=(varepsilon_0h^2)/(pime^2)(n^2)/Z)`
The kinetic energy for the `n^(th)` orbit is `KE_n=1/2 mv_n^2 =(me^4)/(8varepsilon_0^2h^2)(Z^2)/(n^2)`
This implies that `U_n =-2KE_n` . Total energy in the `n^(th)` orbit is
`E_n=KE_n+U_n=KE_n-2KE_n=-KE_n`
`E_n=-(me^4)/(8varepsilon_0^2h^2)(Z^2)/(n^2)`
For hydrogen atom (Z=1) ,
`E_n=-(me^4)/(8varepsilon_0^2h^2)1/(n^2)` joule
Where n stands for principal quantum number . The negative sign in equation (1) indicates that the electron is bound to the nucleus.
Substituting the value of mass and charge of an electron (m and e ), permittivity of free space `varepsilon_0` and Planck.s constant h and expressing in terms of eV, we get
`E_n=-13.6 1/(n^2) eV`
For the first orbit (ground state ) the total energy of electron is `E_1 =-13.6` eV. for the second excited state), the total energy of electron is `E_3 =-1.51 ` eV and so on.
Notice that the energy of the first excited state is and so on. Thus the orbit which is closest to the nucleus `(r_1)` has lowest lowest energy (minimum energy compared with other orbits) . so. it is often called ground state energy (lowest energy state). The ground state energy of hydrogen (-13.6 eV ) is used as a unit of energy called Ryberg (1 Rydberg =-13.6 eV ).
The negative value of this energy is because of the way the zero of the potential energy is defined . When the electron is taken away to an infinite distance (very far distance) from nucleus, both the potential energy and kinetic energy terms vanish and hence the total energy also vanishes.