The potential energy of the orbital electron in the ground state of hydrogen atoms is $-\mathrm{E}$. What is the kinetic energy?

Question

Aicharya · Accepted Answer

The potential energy of $\mathrm{U}(\mathrm{n}=1)=-\mathrm{E}$ Ground state: total energy is given by $\mathrm{TE}=\mathrm{U}+\mathrm{K}$ $\mathrm{U}=-\frac{\left(\mathrm{ze}^2ight)}{4 \pi \varepsilon_0}$ For orbit consider force balance $\frac{\mathrm{ze}^2}{4 \pi \varepsilon_0 \mathrm{r}^2}=\frac{\mathrm{mv}^2}{\mathrm{r}}$ $	herefore \mathrm{K}=\frac{\mathrm{mv}^2}{2}=\frac{1}{2}\left(\frac{\mathrm{ze}^2}{4 \pi \varepsilon_0 \mathrm{r}^2}ight)$ Thus, $\mathrm{K}=\frac{\mathrm{E}}{2}=\frac{\mathrm{U}}{2}$

The potential energy of the orbital electron in the ground state of hydrogen atoms is $-\mathrm{E}$. What is…

Solution