Question

In hydrogen atoms, the electron makes $6.6 \times {10^{15}}\,revolution\,per\,second$ around the nucleus in an orbit of radius $0.5 \times {10^{ - 10}}m$ . It is equivalent to a current nearly:
(A) $1A$
(B) $1mA$
(C) $1\mu A$
(D) $1.6 \times {10^{ - 19}} A$

Answer 1

To conclude the required current, as we know that the time period is inversely proportional to the frequency and the frequency is already given, so we will apply the formula of current in the terms of time period, i.e.. $\therefore i = \dfrac{q}{t}$ .

Formula-used:
Formula of current in which current is directly proportional to the charge of an electron and inversely proportional to the time period, i.e.. $i = \dfrac{q}{t}$ .
And the frequency is inversely proportional to the time period: $f = \dfrac{1}{t}$ .

Complete step by step solution:
In hydrogen atoms, revolution made by the electron per second is the frequency of the electron, i.e.. $f = 6.6 \times {10^{15}}\,$ .
Radius of the nucleus is, $r = 0.5 \times {10^{10}}m$
So, we will apply the formula of current in the terms of charge on an electron and the time period:
$\therefore i = \dfrac{q}{t}$ ……….(i)
where, $i$ is the current,
$q$ is the charge on an electron and,
$t$ is the time period.
As we have already the value of frequency, so we will relates the time period with the frequency:
$\therefore f = \dfrac{1}{t}$ ……(ii)
where, $f$ is the frequency of the electron,
$t$ is the time period.
As we know, the time period is, $t = \dfrac{{2\pi r}}{V}$ , where $r$ is the radius of the nucleus and $V$ is the velocity of the electron around the nucleus.
$\Rightarrow f = \dfrac{1}{{\dfrac{{2\pi r}}{V}}} = \dfrac{V}{{2\pi r}}$
Now, as comparing eq(i) and eq(ii), we get:-
$\because i = q \times f$
Now, as we know that, that charge on an electron is, $q = 1.6 \times {10^{ - 19}}$
So, we will substitute the value of charge and the frequency in the above equation:
$\Rightarrow i = 1.6 \times {10^{ - 19}} \times 6.6 \times {10^{15}}\\\ \Rightarrow i = 1mA \\\$
Therefore, the required current is $1mA$ .
Hence, the correct option is (B) $1mA$ .

Note: The duration of one complete vibration is the time period. The quantities of frequency and period are inversely linked. The number of full waveforms generated each second is the frequency $(f)$ of a wave. The number of oscillations per second is the same as the number of repeats per second.