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Q 41

Physics for Scientists and Engineers: A Strategic Approach with Modern Physics
Found in: Page 1083

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Short Answer

A 222Rn atom (radon) in a 0.75 T magnetic field undergoes radioactive decay, emitting an alpha particle in a direction perpendicular to . The alpha particle begins cyclotron motion with a radius of 45 cm. With what energy, in MeV, was the alpha particle emitted?

The energy of alpha particle emitted is 5.49 MeV

See the step by step solution

Step by Step Solution

Step 1. Given information is :Magnetic field = 0.74 TRadius of initial cyclotron motion = 45 cm

We need to find out the energy of alpha particle emitted.

Step 2. Using the equilibrium condition between centripetal force and magnetic force

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