Question

Constants Periodic Table Part A A doubly charged helium atom whose mass is 6.6 x 10-27 kg is accelerated by a voltage of 3400 V What will be its radius of curvature if it moves in a plane perpendicular to a uniform 0.30 T field? Express your answer using two significant figures. 0四 Submit Request Answer ▼ Part B What is its period of revolution? Express your answer using two significant figures Submit

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 · 3400V / 6.6×10-27kg)
= 57.46*10^4m/s
Since v << c=3×10+8m/s assumption was correct.

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 · 5.746×10^5m/s / (2*1.6022×10-19C · 0.30T)
= 39.45×10^-3m
=0.03945m
(b)
T = 2·π/ω
= 2·π·r/v
= 2 · π · 0.03945m / 5.746×10^5m/s
= 4.311×10-7s