Question

An electron in a hydrogen atom undergoes a transition from an orbit with quantum number ni to another with quantum number nf. Vi and Vf are respectively the initial and final potential energies of the electron. If Vi/ Vf = 6.25, then the smallest possible nf.

Answer 1

The potential energy of an electron in Bohr’s modal is given by (assuming Coulombic force)

U = −Kze² / γ

 

Where, γ = radius

& radius for Bohr’s orbital (for mono-electronic system) =0.529 n² / z

 

so, Ui / Ur = γ2 / γ1= n2² / n1² …(1)

As, given n2 / n1 = 6.25

 

By comparing the above equation and equation (1).

n2 / n1 = √6.25

n2 / n1 = 2.5

So, n2 / n1 = 5 / 2

 

It can be written as,

n2 = 5 n1 = 2

so the final value of n2 = 5.

 

 

The final value = 5