An electron rotates in a circle around a nucleus having positive charge $Ze$. The correct relation between the total energy $(E)$ of the electron and its potential energy $(U)$ is:

  • A
    $E = 2U$
  • B
    $2E = 3U$
  • C
    $E = U$
  • D
    $2E = U$

Explore More

Similar Questions

Minimum excitation potential of Bohr's first orbit in hydrogen atom is.....$V$

Ionization potential of a hydrogen atom is $13.6 \text{ eV}$. Hydrogen atoms in the ground state are excited by monochromatic radiation of photon energy $12.1 \text{ eV}$. According to Bohr's theory,the number of spectral lines emitted by the hydrogen atoms will be:

The inverse square law in electrostatics is $|\vec F| = \frac{{{e^2}}}{{4\pi { \in _0}{r^2}}}$ for the force between an electron and a proton. The $\frac{1}{r^2}$ dependence of $|\vec F|$ can be understood in quantum theory as being due to the fact that the particle of light (photon) is massless. If photons had a mass $m_p$,the force would be modified to $|\vec F| = \frac{{{e^2}}}{{4\pi { \in _0}}}\left( {\frac{1}{{{r^2}}} + \frac{\lambda }{r}} \right)\left( {{e^{ - \lambda r}}} \right)$ where $\lambda = \frac{{{m_p}c}}{\hbar }$ and $\hbar = \frac{h}{{2\pi }}$. Estimate the change in the ground state energy of a $H$-atom if $m_p$ were $10^{-6}$ times the mass of an electron.

Which state of triply ionised beryllium $(Be^{3+})$ has the same orbital radius as that of the ground state of hydrogen?

The ionization energy of hydrogen is $13.6 \, eV$. If $h = 6.6 \times 10^{-34} \, J \cdot s$,the order of magnitude of the Rydberg constant $R$ will be:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo