- #1
superjen
- 26
- 0
A doubly charged helium atom (mass = 6.68 x 10-27 kg) is accelerated through a potential difference of 4.00x 103 V. What will be the radius of curvature of the path of the atom if it is in a uniform 0.450 T magnetic field?
the equation i was using was
r = mV/|q|B
m = 6.68 x 10^-27
V = 4.00 x 10^3
B = 0.450T
q = 1.6 x 10^-19
I don't think this is right. any help pr tips?
Thanks :)
the equation i was using was
r = mV/|q|B
m = 6.68 x 10^-27
V = 4.00 x 10^3
B = 0.450T
q = 1.6 x 10^-19
I don't think this is right. any help pr tips?
Thanks :)