Question

The ground state energy of hydrogen atom is $-13.6\, eV$. What is the potential energy of the electron in this state?

• A. 0 eV
• B. -27.2 eV
• C. 1 eV
• D. 2 eV

Answer 1

Answer: (B)

-27.2 eV

Answer 2

From the first postulate of Bohr's atom model. $\frac{m v^{2}}{r} =\frac{K Z e^{2}}{r^{2}}$ $\Rightarrow \frac{1}{2} m v^{2} =\frac{1}{2} \frac{K Z e^{2}}{r}$ i.e., $KE$ of electron $=\frac{1}{2} m v^{2}=\frac{K Z e^{2}}{2 r}$ ...(1) Potential due to the nucleus, in the orbit in which electron is revolving $=\frac{K Z e}{r}$ $\therefore$ Potential energy of electron potential $\times$ charge $=\frac{K Z e}{r}(-e)=-\frac{K Z e^{2}}{r}$ ...(2) Total energy of electron in the orbit $E = KE + PE$ $=\frac{1}{2} \frac{K Z e^{2}}{r}-\frac{K Z e^{2}}{r}$ $=-\frac{K Z e^{2}}{2 r}$ ...(3) This implies that $PE =2 E$ Given, that in ground state of hydrogen, total energy $E =-13.6\, eV$ $PE =-2 \times 13.6$ Total energy of electron in outer orbits is more than