Question

A proton, an electron, and a Helium nucleus, have the same energy. They are in circular orbits in a plane due to magnetic field perpendicualr to the plane. Let \(r_p\), \(r_e\) and \(r_{He}\) be their respective radii, then

• A. \(r_e \lt r_p \lt r_{He}\)
• B. \(r_e \lt r_p = r_{He}\)
• C. \(r_e \gt r_p \gt r_{He}\)
• D. \(r_e \gt r_p = r_{He}\)

Hint:

For equal kinetic energy K, \(r=\dfrac{\sqrt{2mK}}{qB}\propto\dfrac{\sqrt m}{q}\); compare \(\sqrt m/q\) for the three particles.

Answer 1

The Correct Option is B

In a magnetic field \(qvB=\dfrac{mv^2}{r}\Rightarrow r=\dfrac{mv}{qB}=\dfrac{\sqrt{2mK}}{qB}\). For the same K and B, \(r\propto\dfrac{\sqrt m}{q}\).
Proton: \(\dfrac{\sqrt{m_p}}{e}\)
Electron: \(\dfrac{\sqrt{m_e}}{e}\), and \(m_e\ll m_p\), so \(r_e\lt r_p\).
Helium nucleus: \(m=4m_p,\ q=2e\Rightarrow \dfrac{\sqrt{4m_p}}{2e}=\dfrac{\sqrt{m_p}}{e}\), so \(r_{He}=r_p\).
Hence \(r_e\lt r_p=r_{He}\).
Answer: (2) \(r_e \lt r_p = r_{He}\)