Question

The total energy of an electron in the first excited state of hydrogen is about -3.4eV. Its kinetic energy in this state is:

• A. - 3.4 eV
• B. - 1.7 eV
• C. 1.7 eV
• D. 3.4 eV

Answer 1

Answer: (D)

3.4 eV

Answer 2

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