Question:
A deuteron and an alpha particle having equal kinetic energy enter perpendicularly into a magnetic field. Let \(r_d\) and \(r_\alpha\) be their respective radii of circular path. The value of \(\frac{r_d}{r_\alpha}\) is equal to :
A deuteron and an alpha particle having equal kinetic energy enter perpendicularly into a magnetic field. Let \(r_d\) and \(r_\alpha\) be their respective radii of circular path. The value of \(\frac{r_d}{r_\alpha}\) is equal to :
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For a charged particle in a magnetic field \(r = \frac{\sqrt{2mK}}{qB}\), so at equal K, \(r \propto \frac{\sqrt{m}}{q}\).
Updated On: Nov 4, 2026
- 1
- 2
- \(\sqrt{2}\)
- \(\frac{1}{\sqrt{2}}\)
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The Correct Option is C
Solution and Explanation
Radius of the circular path: \(r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{qB}\).
For equal kinetic energy K and the same B, \(r \propto \frac{\sqrt{m}}{q}\).
Deuteron: mass \(= 2u\), charge \(= e\). Alpha particle: mass \(= 4u\), charge \(= 2e\).
\(\frac{r_d}{r_\alpha} = \frac{\sqrt{2}/1}{\sqrt{4}/2} = \frac{\sqrt2}{1} = \sqrt{2}\).
Answer: (3) \(\sqrt{2}\)
For equal kinetic energy K and the same B, \(r \propto \frac{\sqrt{m}}{q}\).
Deuteron: mass \(= 2u\), charge \(= e\). Alpha particle: mass \(= 4u\), charge \(= 2e\).
\(\frac{r_d}{r_\alpha} = \frac{\sqrt{2}/1}{\sqrt{4}/2} = \frac{\sqrt2}{1} = \sqrt{2}\).
Answer: (3) \(\sqrt{2}\)
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