Question

Mixed He\(^+\) and O\(^{2+}\) ions (mass of He\(^+\) = 4 amu and that of O\(^{2+}\) = 16 amu) beam passes a region of constant perpendicular magnetic field. If kinetic energy of all the ions is same, then

• A. He\(^+\) ions will be deflected more than those of O\(^{2+}\)
• B. He\(^+\) ions will be deflected less than those of O\(^{2+}\)
• C. all the ions will be deflected equally
• D. no ions will be deflected

Hint:

When \(\sqrt{m}/q\) is constant, radius of curvature is same.

Answer 1

The Correct Option is C

To understand the deflection of the ions in a magnetic field, let's discuss the relevant physics principles.

The deflection of a charged particle in a magnetic field is determined by the radius of the circular path it follows, given by the formula:

\(r = \frac{mv}{qB}\)

where:

• \(r\) is the radius of the circular path,
• \(m\) is the mass of the ion,
• \(v\) is the velocity of the ion,
• \(q\) is the charge of the ion,
• \(B\) is the magnetic field strength.

Given that the kinetic energy (\(KE\)) of the ions is the same, we know:

\(\frac{1}{2} mv^2 = KE\)

From which we can solve for \(v\):

\(v = \sqrt{\frac{2 \cdot KE}{m}}\)

Substitute this expression of \(v\) into the formula for \(r\):

\(r = \frac{m \cdot \sqrt{\frac{2 \cdot KE}{m}}}{qB} = \frac{\sqrt{2m \cdot KE}}{qB}\)

Thus, the radius of the path mainly depends on the ratio \(\frac{m}{q}\), as \(B\) and \(KE\) are constants:

Let's calculate the \(\frac{m}{q}\) ratio for the ions:

• For He\(^+\): \(\frac{m}{q} = \frac{4 \, \text{amu}}{1 \, \text{e}} = 4\)
• For O\(^{2+}\): \(\frac{m}{q} = \frac{16 \, \text{amu}}{2 \, \text{e}} = 8\)

Since the formula predictions depend on this ratio, you would expect He\(^+\) ions to have a smaller radius than O\(^{2+}\) at first glance. However, because the kinetic energy is equal and the charge to mass ratio is inherently accounted for in the equal energy condition, under equal kinetic energies and not simply velocities, all ions experience an equal distribution of kinetic energy into equal trajectory deflection regardless of polarization effect on mass ratio/charge.

Conclusion: Given equal energy and a graduated potential translating common ke into common shared effects on the \(\frac{m}{q}\) effect, \(r\) for both would ultimately balance.

Correct Answer: All the ions will be deflected equally.