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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  VfVi=6.25, then the smallest possible nf is


Solution

Potential energy of a electron in Bohr's modal is given by (assuming Coulombic force) U=Kze2γ
where, γ= radious 
& radius for Bohr's orbital (for mono-electronic system) =0.529n2z
so, UiUr=γ2γ1=n22n21 ...(1)

As, given n2n1=6.25
by comparing above equation to equation (1).
 n2n1=6.25n2n1=2.5

So, n2n1=52
it can be written as, 
n2=5, n1=2
so the final value of n2=5.


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