Question

a helium atom, with a charge of 2e and a mass of 6.6 × 10 to the -27 kg, is accelerated by a voltage of 3200 Volts in a uniform magnetic field of 0.240 T. A) what is the speed of the helium atom? B) what is the radius of curvature of the helium atom? C) what is the helium atom's period of revolution?

Answer 1

(a)
First calculate velocity from energy balance:
Assume negligible relativistic effects
ΔE_kin = ΔE_el
<=>
(1/2)·m·v² = q·U
=>
v = √(2·q·E/m)
= √(2 * 2*1.6022×10-19C * 3200V / 6.6×10-27kg)
= 5.5743×10^5m/s


b) For stationary motion Lorentz force equals centripetal force on the moving atom
m·ω²·r² = q·v·B
<=>
m·v²/r = q·v·B
=>
r = m·v / (q·B)
= 6.6×10^-27kg * 5.574×10^5m/s / (2*1.6022×10-19C * 0.240T)
= 0.04783m

(b)
T = 2·π/ω
= 2·π·r/v
= 2 * π * 0.04783m / 5.574×10^5m/s
= 5.3915×10-7s