A) 5.1 V
B) 12.1 V
C) 17.2 V
D) 7 V
Correct Answer: D
Solution :
For \[n=1,{{E}_{1}}=-\frac{13.6}{{{(1)}^{2}}}=-13.6\,eV\] and for \[n=3,{{E}_{3}}=-\frac{13.6}{{{(3)}^{2}}}=-1.51\,eV\] So, required energy \[E={{E}_{3}}-{{E}_{1}}=-(1.51)-(-13.6)\] \[=12.09\,eV\] \[\because \] \[E=W+eV\] \[\therefore \] \[eV=E-W\] \[eV=(12.09-5.1)e\] \[V=7\,volt\]You need to login to perform this action.
You will be redirected in
3 sec