The total energy of an electron in the first excited state of hydrogen is about -3.4eV. Its kinetic energy in this state is:
- 3.4 eV
- 1.7 eV
1.7 eV
3.4 eV
The Correct Option is D
Solution and Explanation
Kinetic energy of electron is, KE=\(\frac{13.6Z2}{n^2}\) eV
For the first excited state of the hydrogen atom, n=2 and Z=1
∴ KE = \(\frac{13.6}{2^2}\) =3.4 eV
Total energy , E=KE+PE
\(\Rightarrow\) −3.4 = 3.4+PE
∴ PE=−6.8 eV
Therefore, the correct option is (D): 3.4 eV
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