Question

Assuming the atom is in the ground state, the expression for the magnetic field at a point nucleus in hydrogen atom due to circular motion of electron is (\(\mu _{0}\) = permeability of free space, \(\epsilon _{0}\) = permittivity of free space, \(m\) = mass of electron, \(e\) = electronic charge)

• A. \(\frac{\mu _{0}e^{2}m^{2}}{8\epsilon _{0}h^{2}}\)
• B. \(\frac{\mu _{0}e^{7}\pi m^{2}}{8\epsilon _{0}^{3}h^{5}}\)
• C. \(\frac{\mu _{0}e^{5}\pi m}{6\epsilon _{0}^{2}h^{2}}\)
• D. \(\frac{\mu _{0}e^{2}\pi m^{2}}{8\epsilon _{0}h^{2}}\)

Hint:

Treat the orbiting electron as a circular current loop and use Bohr's ground state radius and speed.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The orbiting electron acts as a current loop. The field at its centre is \(B=\dfrac{\mu_0 I}{2r}\) with \(I=\dfrac{ev}{2\pi r}\).

Step 2: Combine
\(B=\dfrac{\mu_0 e v}{4\pi r^2}\)

Step 3: Ground State Values
Bohr's theory gives \(r=\dfrac{\epsilon_0h^2}{\pi me^2}\) and \(v=\dfrac{e^2}{2\epsilon_0h}\).

Step 4: Substitute
\(B=\dfrac{\mu_0e}{4\pi}\cdot\dfrac{e^2}{2\epsilon_0h}\cdot\dfrac{\pi^2m^2e^4}{\epsilon_0^2h^4}\)
\(B=\dfrac{\mu_0e^7\pi m^2}{8\epsilon_0^3h^5}\), which is option (B).

Final Answer:
The magnetic field at the nucleus is \(\dfrac{\mu_0e^7\pi m^2}{8\epsilon_0^3h^5}\). \[ \boxed{\text{(B) }\dfrac{\mu _{0}e^{7}\pi m^{2}}{8\epsilon _{0}^{3}h^{5}}} \]