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The average current due to an electron o...

The average current due to an electron orbiting the proton in the `n^(th)` Bohr orbit of the hydrogen atom is

A

`propn`

B

`propn^(3)`

C

`propn^(-3)`

D

none of these

Text Solution

AI Generated Solution

The correct Answer is:
To find the average current due to an electron orbiting the proton in the \( n^{th} \) Bohr orbit of the hydrogen atom, we can follow these steps: ### Step 1: Understand the relationship between radius and quantum number The radius of the \( n^{th} \) Bohr orbit is given by the formula: \[ R_n \propto \frac{n^2}{Z} \] For hydrogen, \( Z = 1 \), so we can simplify this to: \[ R_n \propto n^2 \] ### Step 2: Determine the velocity of the electron in the orbit The velocity \( V \) of the electron in the \( n^{th} \) orbit is given by: \[ V_n \propto \frac{Z}{n} \] Again, for hydrogen, this simplifies to: \[ V_n \propto \frac{1}{n} \] ### Step 3: Calculate the angular frequency \( \omega \) The angular frequency \( \omega \) is defined as: \[ \omega = \frac{V}{R} \] Substituting the expressions for \( V \) and \( R \): \[ \omega \propto \frac{Z/n}{n^2/Z} = \frac{Z^2}{n^3} \] For hydrogen, this simplifies to: \[ \omega \propto \frac{1}{n^3} \] ### Step 4: Relate current to charge and time The average current \( I \) due to the orbiting electron can be expressed as: \[ I = \frac{q}{T} \] Where \( q \) is the charge of the electron (denoted as \( e \)) and \( T \) is the time period of one complete revolution. The time period \( T \) can be expressed in terms of angular frequency: \[ T = \frac{2\pi}{\omega} \] Substituting this into the current formula gives: \[ I = e \cdot \omega \cdot \frac{1}{2\pi} \] ### Step 5: Substitute for \( \omega \) Substituting the expression for \( \omega \): \[ I \propto e \cdot \frac{Z^2}{n^3} \cdot \frac{1}{2\pi} \] For hydrogen, this simplifies to: \[ I \propto \frac{1}{n^3} \] ### Conclusion Thus, the average current due to an electron orbiting the proton in the \( n^{th} \) Bohr orbit of hydrogen is: \[ I \propto n^{-3} \] ### Final Answer The average current is inversely proportional to the cube of the principal quantum number \( n \): \[ I \propto n^{-3} \] ---
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