© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Write down the expression for the radius of the circular path
When a charged particle of mass $m$ and charge $q$ moves in a uniform magnetic field $B$ (perpendicular to its velocity), it follows a circular path whose radius $r$ is given by:
$r = \frac{mv}{qB}$
Step 2: Relate the velocity to the same kinetic energy
The problem states that the proton, electron, and Helium nucleus all have the same kinetic energy $K$. Since the kinetic energy $K = \frac{1}{2} m v^2$, the speed $v$ can be expressed as:
$v = \sqrt{\frac{2K}{m}}$
Substituting this into the radius expression gives:
$r = \frac{m \sqrt{\frac{2K}{m}}}{qB} = \frac{\sqrt{2mK}}{qB}$
This shows that, for the same kinetic energy $K$ and the same magnetic field $B$, the radius depends on $\sqrt{m}$ divided by $q$.
Step 3: Compare the particles
• Electron ($e^-$): Mass $m_e$, charge $-\,e$ (magnitude $e$)
• Proton ($p^+$): Mass $m_p$, charge $+\,e$
• Helium nucleus ($\alpha$ or $He^{2+}$): Mass $\approx 4m_p$, charge $+\,2e$
Hence, for the same kinetic energy $K$:
Electron radius:
$r_e \propto \frac{\sqrt{m_e}}{e}$
Proton radius:
$r_p \propto \frac{\sqrt{m_p}}{e}$
Helium nucleus radius:
$r_{He} \propto \frac{\sqrt{4m_p}}{2e} = \frac{2\sqrt{m_p}}{2e} = \frac{\sqrt{m_p}}{e}$
We see that $r_p = r_{He}$, because both have the same proportionality factor. Since $m_e \ll m_p$, we get $r_e < r_p = r_{He}$.
Step 4: State the conclusion
The ordering of the radii is:
$r_e < r_p = r_{He}$
This matches the correct answer given as: $r_e < r_p = r_{He}$.
