A) 3.4 eV
B) 6.8 eV
C) 13.6 eV
D) 1.7 eV
Correct Answer: A
Solution :
Total energy of electron in the orbit is equal to negative of its kinetic energy. The energy of hydrogen atom when the electron revolves in nth orbit is \[E=\frac{-13.6}{{{n}^{2}}}eV\] In the ground state\[n=1\] \[E=\frac{-13.6}{{{1}^{2}}}=-13.6\,eV\] For \[n=2,E=\frac{-13.6}{{{2}^{2}}}=-3\,eV\] So, kinetic energy of electron in the first excited state (i.e., for\[n=2\]) is \[K=-E=-(-3.4)=3.4\,eV\]
You need to login to perform this action.
You will be redirected in
3 sec
