A proton and an `alpha`-particle, accelerated through same potential difference, enter a region of uniform magnetic field with their velocities perpendicular to the field. Compare the radii of circular paths followed by them.

A proton and an `alpha`-particle, accelerated through same potential d

1.	A proton and an `alpha`-particle, accelerated through same potential difference, enter a region of uniform magnetic field with their velocities perpendicular to the field. Compare the radii of circular paths followed by them.
Answer» Let mass of proton = m, charge of proton = e Now, Mass of `alpha`-particle = 4m Chanrge of `alpha`-particle = 2e When a charge q is accelerated by V volts, it acquires a kinetic energy `E_(k) = qV` `:.` Momentum is given by `mv = sqrt(2m E_(k)) = sqrt(2mqV)` Radius `r = (mv)/(qB) or r = (sqrt(2mqV))/(qB) = sqrt((2mV)/(qB^(2))` Thus `(r_(p))/(r_(alpha)) = sqrrt((2mV)/(eB^(2))) xx sqrt((2eB^(2))/(2(4m) V)) = (1)/(sqrt2)`

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A proton and an `alpha`-particle, accelerated through same potential difference, enter a region of uniform magnetic field with their velocities perpendicular to the field. Compare the radii of circular paths followed by them.

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