Question:
A deuteron of kinetic energy 50keV is describing a circular orbit of radius 0.5m in a plane perpendicular to the magnetic field B. The kinetic energy of the proton that describes a circular orbit of radius 0.5m in the same plane with the same B is:
A deuteron of kinetic energy 50keV is describing a circular orbit of radius 0.5m in a plane perpendicular to the magnetic field B. The kinetic energy of the proton that describes a circular orbit of radius 0.5m in the same plane with the same B is:
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For same radius in magnetic field: K ∝ (1)/(m).
Updated On: Mar 20, 2026
- \(25\,\text{keV}\)
- \(50\,\text{keV}\)
- \(100\,\text{keV}\)
- 200keV
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Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Radius of circular motion: r = (mv)/(qB)
Step 2: For same r and B: √(mK) = constant
Step 3: Since md = 2mp: Kp = 2Kd = 100keV
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