An alpha particle ( q = +2 e , m = 4.00u) travels in a circular path of radius 4

Question

An alpha particle (q = +2e, m = 4.00u) travels in a circular path of radius 4.88 cm in a uniformmagnetic field with B = 1.00 T. Calculate(a) its speed, (b) its period ofrevolution, (c) its kinetic energy (in eV), and(d) the potential difference (in V) through whichit would have to be accelerated to achieve this energy.

(a)

Number

Units

(b)

Number

Units

(c)

Number

Units

(d)

Number

Units

(a)

Number

Units

Explanation / Answer

(c) KE = 0.5mv2 = 0.5 * 4 * ( 2.3377 x106 )2 = 10.92 M J (d) We know that            KE = q V       ==> V = KE / q                 = 10.92 x 1012 / 3.2 x 10-19                 = 3.4155 x 1031 V

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