In any Bohr orbit of a hydrogen atom,the ratio of $K.E.$ to $P.E.$ of a revolving electron at a distance $r$ from the nucleus is:

  • A
    $-1$
  • B
    $+\frac{1}{2}$
  • C
    $1$
  • D
    $-\frac{1}{2}$

Explore More

Similar Questions

Consider a hydrogen-like ionized atom with atomic number $Z$ with a single electron. In the emission spectrum of this atom,the photon emitted in the $n = 2$ to $n = 1$ transition has energy $74.8 \ eV$ higher than the photon emitted in the $n = 3$ to $n = 2$ transition. The ionization energy of the hydrogen atom is $13.6 \ eV$. The value of $Z$ is:

The following parameter is the same for all hydrogen-like atoms and ions in their ground state.

If the radius of the first Bohr orbit is $r$, then the radius of the second Bohr orbit will be

For a certain hypothetical one-electron atom,the wavelength (in $\mathring{A}$) for the spectral lines for transition from $n = p$ to $n = 1$ is given by $\lambda = \frac{1500p^2}{p^2 - 1}$ (where $p > 1$). The ionization potential of this element must be .....$V$ (Take $hc = 12420\ eV\cdot\mathring{A}$).

Difficult
View Solution

The triply ionized beryllium $(Be^{3+})$ has the same electron orbital radius as that of the ground state of hydrogen. Hence, the energy state of triply ionized beryllium is (Given $Z = 4$ for beryllium)

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