An alpha particle (charge + 2e) travels in a circular path of radius. 5m in a magnetic field of 1.0 T. Find the (a) period, (b) speed, and (c) kinetic energy (in electron volts) of the alpha particle. Take m = 6.65 x 10 - 27 kg for the mass of the alpha particle.

Given Information:

Radius = r = 0.5 m

Magnetic field = 1.0 T

Required Information:

Period = T = ?

Speed = v = ?

Kinetic energy = KE = ?

Answer:

Period = 0.13x10⁻⁶ seconds

speed = 24.16x10⁶ m/s

Kinetic energy = 12.11 MeV

Explanation:

(a) period

The time period of alpha particle is related to its orbital speed as

T = 2πr/v eq. 1

According to newton's law

F = ma

Force due to magnetic field is given by

F = qvB

qvB = ma

qvB = m (v²/r)

qB = mv/r

v = qBr/m eq. 2

substitute the eq. 2 in eq. 1

T = 2πr/qBr/m

r cancels out

T = 2π/qB/m

T = 2πm/qB

T = 2π*6.65x10⁻²⁷/2*1.602x10⁻¹⁹*1

T = 0.13x10⁻⁶ seconds

(b) speed

From equation 1

T = 2πr/v

v = 2πr/T

v = 2π*0.5/0.13x10⁻⁶

v = 24.16x10⁶ m/s

(c) kinetic energy (in electron volts)

Kinetic energy is given by

KE = 0.5mv²

KE = 0.5*6.65x10⁻²⁷ * (24.16x10⁶) ²

KE = 1.94x10⁻¹² J

since 1 electron volt has 1.602x10⁻¹⁹ J

KE = 1.94x10⁻¹²/1.602x10⁻¹⁹

KE = 12.11 MeV