JEE Advance - Physics (2017 - Paper 1 Offline - No. 8)
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
Answer
5
Explanation
The energy of a hydrogen atom with an electron in nth orbit is given by
$${E_n} = - {{13.6} \over {{n^2}}}eV$$.
The potential energy in nth orbit is related to En by Vn = En/2. Thus, potential energies in orbits with quantum numbers ni and nf 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 nf = 2.5ni. Since ni and nf are positive integers, smallest possible integral value of nf is 5 for ni = 2.
Comments (0)
