Question
Determine the maximum angular speed of an electron moving in a stable orbit around the nucleus of hydrogen atom.
✓
Answer
The radius of the ,ith Bohr orbit is
$r=\frac{\varepsilon_0 h^2 n^2}{\pi m Z e^2}$
and the linear speed of an electron in this orbit is
$v =\frac{Z e^2}{2 \varepsilon_0 n h}$
where $\varepsilon_0$ permittivity of free space, $h \equiv$ Planck's constant, $n \equiv$ principal quantum number, $m \equiv$ electron mass, e electronic charge and $Z \equiv$ atomic number of the atom.
Since angular speed $\omega=\frac{v}{r}$, then from Eqs. (1) and (2), we get,
$\omega=\frac{v}{r}=\frac{Z e^2}{2 \varepsilon_0 n h} \cdot \frac{\pi m Z e^2}{\varepsilon_0 h^2 n^2}=\frac{\pi m Z^2 e^4}{2 \varepsilon_0^2 h^3 n^3}$
which gives the required expression for the angular speed of an electron in the nth Bohr orbit.
From Eq. (3), the frequency of revolution of the electron,
$f =\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \times \frac{\pi m Z^2 e^4}{2 \varepsilon_0^2 h^3 n^3}=\frac{m Z^2 e^4}{4 \varepsilon_0^2 h^3 n^3}$
as required.
[Note: From Eq. (4), the period of revolution of the electron, $T =\frac{1}{f}=\frac{4 \varepsilon_0^2 h^3 n^3}{m Z e^4}$. Hence, $f \propto$ $\frac{1}{n^3}$ and $\left.T \propto n^3\right]$
Obtain the formula for $\omega$ and continue as follows :
$ \omega(\text { maximum) }) \frac{\pi m e^4}{2 \varepsilon_0^2 h^3} \quad(\text { for } Z =1 \text { and } n=1)$
$=\frac{(3.142)\left(9.1 \times 10^{-31} kg \right)\left(1.6 \times 10^{-19} C \right)^4}{(2)\left(8.85 \times 10^{-12} C ^2 / N \cdot m ^2\right)^2\left(6.63 \times 10^{-34} J \cdot s \right)^3} rad / s $
$ =\frac{(3.142)(9.1)(1.6)^4\left(10^{-107}\right)}{(2)(8.85)^2(6.63)^3\left(10^{-126}\right)}$
$=4.105 \times 10^{16} rad / s $
This is required quantity.
$r=\frac{\varepsilon_0 h^2 n^2}{\pi m Z e^2}$
and the linear speed of an electron in this orbit is
$v =\frac{Z e^2}{2 \varepsilon_0 n h}$
where $\varepsilon_0$ permittivity of free space, $h \equiv$ Planck's constant, $n \equiv$ principal quantum number, $m \equiv$ electron mass, e electronic charge and $Z \equiv$ atomic number of the atom.
Since angular speed $\omega=\frac{v}{r}$, then from Eqs. (1) and (2), we get,
$\omega=\frac{v}{r}=\frac{Z e^2}{2 \varepsilon_0 n h} \cdot \frac{\pi m Z e^2}{\varepsilon_0 h^2 n^2}=\frac{\pi m Z^2 e^4}{2 \varepsilon_0^2 h^3 n^3}$
which gives the required expression for the angular speed of an electron in the nth Bohr orbit.
From Eq. (3), the frequency of revolution of the electron,
$f =\frac{\omega}{2 \pi}=\frac{1}{2 \pi} \times \frac{\pi m Z^2 e^4}{2 \varepsilon_0^2 h^3 n^3}=\frac{m Z^2 e^4}{4 \varepsilon_0^2 h^3 n^3}$
as required.
[Note: From Eq. (4), the period of revolution of the electron, $T =\frac{1}{f}=\frac{4 \varepsilon_0^2 h^3 n^3}{m Z e^4}$. Hence, $f \propto$ $\frac{1}{n^3}$ and $\left.T \propto n^3\right]$
Obtain the formula for $\omega$ and continue as follows :
$ \omega(\text { maximum) }) \frac{\pi m e^4}{2 \varepsilon_0^2 h^3} \quad(\text { for } Z =1 \text { and } n=1)$
$=\frac{(3.142)\left(9.1 \times 10^{-31} kg \right)\left(1.6 \times 10^{-19} C \right)^4}{(2)\left(8.85 \times 10^{-12} C ^2 / N \cdot m ^2\right)^2\left(6.63 \times 10^{-34} J \cdot s \right)^3} rad / s $
$ =\frac{(3.142)(9.1)(1.6)^4\left(10^{-107}\right)}{(2)(8.85)^2(6.63)^3\left(10^{-126}\right)}$
$=4.105 \times 10^{16} rad / s $
This is required quantity.
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