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A deuteron of kinetic energy 50 keV is describing a circular orbit of radius 0.5 metre in a plane perpendicular to the magnetic field B. The kinetic energy of the proton that describes a circular orbit of radius 0.5 metre in the same plane with the same B is
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- 25 keV
- 50 keV
- 200 keV
- 100 keV
Correct Option: D
For a charged particle orbiting in a circular path in a magnetic field
| = Bqv ⇒ v = | ||||
| r | m |
or, mv² = Bqvr
Also,
| EK = | mv² = | Bqvr = Bq. | . | = | |||||
| 2 | 2 | 2 | m | 2m |
| For deuteron, E1 = | ||
| 2 × 2m |
| For proton, E2 = | ||
| 2m |
| = | = | ⇒ | = | ⇒ E2 = 100 keV | ||||
| E2 | 2 | E2 | 2 |
