Question

show that the ground state energy of electrons in hydrogen atom is equal to the first excited state energy of electron in He+ ion assuming rydberg constants are equal

Answer 1

from Bohr's model,  the angular momentum of an electron in an orbit is an integral multiple of h/2π. Mathematically,

mvr=nh/2π

.  (1)

From this equation, you can figure out what the velocity and the radius of this electron would be in that particular orbit.

Now, the centripetal force the electron experiences has to be supplied by the Coulomb interaction between the nucleus and the electron - there's absolutely no other way out! Once again, mathematically, this is:

mv2/r=ke2/r2

  (2)

Solving this equation, you can calculate the Kinetic Energy of the electron as

K=1/2mv2=ke2/2r

 (3)

The potential energy of an electron is given as

P=−ke2/r

 (4)

Thus, the total energy of the electron E=K+P=−ke2/2r

(5)

Substitute the expression for velocity from (1) in equation (5) and solving for radius, r, we get: r=n2h2/k4π2e2m

Now if you consider the ground state, then n = 1, and we can plug the values of all the other constants to get r = 0.05259nm

Plug in this value in the total energy expression given in equation (5) and you have -13.6 eV.

This is the simple explanation.

Quantum mechanically you have, Hψ = Eψ , where E is the electron Binding Energy in a Hydrogen Atom   (6)

The value of E =−(me4/n28π2ε2oh2)ψ

  (7)

However, me4/n28π2h2

is the Rydberg constant (Rh

) which has a value of 2.18 x 10^-18 J

Thus E=−Rh/n2

For the ground state, n = 1 and you get the value of E as -2.18 x 10^ -18J, which when you convert to eV gives you -13.6 eV.

Now to understanding the energy levels:

When you plug in different values of n starting from 1, then 2, 3, 4,... you see that the the Hydrogen atom has now several discreet energy levels which are all a factor of -13.6eV, the lowest energy it can possess. This n is the Principle Quantum number of the H-atom.

Another way to look at it is from the perspective of Binding Energy. An electron in a H-atom can possess several possible binding energies with the lowest being -13.6eV. But you need to understand these energies are quantized - they are discreet since n can only be a whole number. Thus, instead of having a continuum of binding energies, what you actually have is a discreet set of energy levels.

This energy level diagram is basically a result of solving Schrodinger's equation for a single electron H-atom to give you the energy eigen values and the equation          E(n) = -13.6/n^2 gives you the energies of all the allowed levels in the H-Atom.

Answer 2

Explanation:

Energy of the nth orbit in hydrogen like species:

$E_n=R_H\times \frac{Z^2}{n^2}$

where,

$E_n$ = energy of $n^{th}$ orbit

n = number of orbit

Z = atomic number

$R_H$ = Rydberg constant

Ground state energy of electrons in hydrogen atom;

n = 1, Z = 1

$E_{1,H}=R_H\times \frac{1^2}{1^2}=R_H$...[1]

First excited state energy of electron in $He+$ or helium  ion

n = 2, Z = 2

$E_{2,He}=R_h\times \frac{2^2}{2^2}=$...[2]

$E_{1,H}=E_{2,He} =R_H$