Question

The electron of an H-atom is revolving around the nucleus in circular orbit having radius $\dfrac{{{h^2}}}{{4\pi m{e^2}}}$​ with $\dfrac{{2\pi {e^2}}}{h}$ The current produced due to the motion of electron is
A) $\dfrac{{2\pi {m^2}{e^2}}}{{3{h^2}}}$
B) zero
C) $\dfrac{{2{\pi ^2}me}}{{{h^2}}}$
D) $\dfrac{{4{\pi ^2}m{e^5}}}{{{h^3}}}$

Answer 1

Hint
As the radius of H-atom in cgs unit is given as $r = \dfrac{{{h^2}}}{{4\pi m{e^2}}}$and velocity of electron is also given then find the time period of revolution by using formula $T = \dfrac{{2\pi r}}{v}$,after this we know that current is $I = \dfrac{Q}{T}$
On substituting the values, we get the required current.

Complete answer:
As it is given that, The radius of the nucleus in cgs unit is $r = \dfrac{{{h^2}}}{{4\pi m{e^2}}}$
And the velocity of electron is $v = \dfrac{{2\pi {e^2}}}{h}$
Now, the time period of the revolution can be calculated by using the formula
$ \Rightarrow T = \dfrac{{2\pi r}}{v}$
Where, r is the velocity of the electron and v is the velocity of the electron.
Now, substitute the value of r and v in above equation of time period, we get
$ \Rightarrow T = \dfrac{{2\pi r}}{v} = \dfrac{{{h^3}}}{{4{\pi ^2}m{e^4}}}$ ………………… (1)
Now, as we know that current is defined as the rate of change of charge with respect to time I.e.
$I = \dfrac{Q}{T}$ ……………………….. (2)
 The charge on electron is $Q = e$ ……………….. (3)
Put the values of equation (2) and (3) in equation (1), we get
$I = \dfrac{e}{T} = \dfrac{{4\pi m{e^5}}}{{{h^3}}}$
This is the required value of current due to the motion of electrons.
Hence, (D) option is correct.

Note
The time period is defined as the time by the electron for one complete cycle. Frequency of the wave increases when time period decreases. As the current is defined as the rate of change of charge with respect to time i.e. $I = \dfrac{Q}{T}$, it can also be written in terms of frequency i.e. $I = Q \times \nu $.