When the electron orbiting in a hydrogen atom in its ground state moves to the third excited state, the de-Broglie wavelength associated with it
- Awill decrease
- Bwill increase
- Cwill remain same
- Dwill be zero
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Match the following List-$I$ with List-$II$ in connection with Bohr's atomic model.
$A$. Speed of revolution of electron $i$. $\frac{1}{4 \pi \varepsilon_0} \frac{2 \pi Z e^2}{n h}$ $B$. Kinetic energy $ii$. $-\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$ $C$. Total energy $iii$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$ $D$. Frequency $iv$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{4 \pi^2 Z^2 e^4 m}{n^3 h^3}$
| $A$. Speed of revolution of electron | $i$. $\frac{1}{4 \pi \varepsilon_0} \frac{2 \pi Z e^2}{n h}$ |
| $B$. Kinetic energy | $ii$. $-\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$ |
| $C$. Total energy | $iii$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{2 \pi^2 m e^4 Z^2}{n^2 h^2}$ |
| $D$. Frequency | $iv$. $\left(\frac{1}{4 \pi \varepsilon_0}\right)^2 \frac{4 \pi^2 Z^2 e^4 m}{n^3 h^3}$ |
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:
DifficultJEE MAIN 2024
View SolutionThe de-Broglie wavelength of an electron moving in the $n^{\text{th}}$ Bohr orbit of radius $r$ is
The speed of an electron in the orbit of a hydrogen atom in the ground state is
Easy
View SolutionThe magnitude of acceleration of the electron in the $n$th orbit of a hydrogen atom is $a_{H}$ and that of a singly ionised helium atom is $a_{He}$. The ratio of $a_{H} : a_{He}$ is
DifficultKVPY 2017
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