Question 112: Atoms & Nuclei

Question

An electron in a hydrogen atom undergoes a transition from an orbit with quantum number $${n_i}$$ to another with quantum number $${n_f}$$. $${V_i}$$ and $${V_f}$$ are respectively the initial and final potential energies of the electron. If $${{{V_i}} \over {{V_f}}} = 6.25$$, then the smallest possible $${n_f}$$ is

Accepted Answer

The energy of a hydrogen atom with an electron in n
th
orbit is given by
$${E_n} =  - {{13.6} \over {{n^2}}}eV$$.
The potential energy in n
th
orbit is related to E
n
by V
n
= E
n
/2. Thus, potential energies in orbits with quantum numbers n
i
and n
f
are given by
$${V_i} =  - {{13.6} \over {2n_i^2}}eV$$, and $${V_f} =  - {{13.6} \over {2n_f^2}}eV$$, which gives
$${{{V_i}} \over {{V_f}}} = {{n_f^2} \over {n_i^2}} = 6.25$$ (given).
Take square root to get n
f
= 2.5n
i
. Since n
i
and n
f
are positive integers, smallest possible integral value of n
f
is 5 for n
i
= 2.

Atoms & Nuclei

An electron in a hydrogen atom undergoes a transition from an orbit with quantum number ${n_i} to another with quantum number {n_f}. {V_i} and {V_f} are respectively the initial and final potential energies of the electron. If {{{V_i}} \over {{V_f}}} = 6.25, then the smallest possible {n_f}$ is

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