The ratio of the speed of an electron in the ground state of the Bohr's first orbit of a hydrogen atom to the velocity of light in air is:
- A$\frac{e^2}{2\varepsilon_0 hc}$
- B$\frac{2e^2\varepsilon_0}{hc}$
- C$\frac{e^3}{2\varepsilon_0 hc}$
- D$\frac{2\varepsilon_0 hc}{e^2}$
Explore More
Similar Questions
The energy of $He^{+}$ ion in its first excited state is $......eV$ (The ground state energy for the Hydrogen atom is $-13.6\,eV$).
What is ionisation energy?
Easy
View SolutionThe acceleration of an electron in the first orbit of the hydrogen atom $(Z = 1)$ is
DifficultJEE MAIN 2017
View SolutionMagnetic moment due to the motion of the electron in the $n$th energy state of a hydrogen atom is proportional to ..........
$A$ diatomic molecule is made of two masses $m_1$ and $m_2$ separated by a distance $r$. Applying the Bohr's quantization rule for angular momentum,calculate its rotational kinetic energy. It is given by the formula:
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo