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

Question

An electron in a hydrogen atom undergoes a transition from an orbit with quantum number ði to another with quantum number ðf. ði and ðf are respectively the initial and final potential energies of the electron. If (Vi/Vf)=6.25, then the smallest possible ðf is (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

| V | propto (1/ n 2) (| V i |/| V f |)=( n f 2/ n i 2)=6.25 ( n f 2/ n i 2)=6.25 ( n f / n i )=√6.25=2.5 n f =2.5 n i Smallest value of ni=2 for which nf=5

Q. 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 Vf​Vi​​=6.25, then the smallest possible nf​ is

∣V∣∝n21​ ∣Vf​∣∣Vi​∣​=ni2​nf2​​=6.25 ni2​nf2​​=6.25 ni​nf​​=6.25​=2.5 nf​=2.5ni​ Smallest value of ni​=2 for which nf​=5

$∣ V ∣ \propto \frac{1}{n ^{2}}$
$\frac{∣ V _{i} ∣}{∣ V _{f} ∣} = \frac{n _{f}^{2}}{n _{i}^{2}} = 6.25$
$\frac{n _{f}^{2}}{n _{i}^{2}} = 6.25$
$\frac{n _{f}}{n _{i}} = 6.25 = 2.5$
$n_{f} = 2.5 n_{i}$
Smallest value of $n_{i} = 2$ for which $n_{f} = 5$