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 |
\[E={{E}_{3}}-{{E}_{1}}=-(1.51)-(-13.6)\] |
\[=12.09\,eV\] |
\[\because \] \[E=W+eV\] |
\[\therefore \] \[eV=E-W\] |
\[eV=(12.09-5.1)e\] |
\[V=7\,volt\] |
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