The most common isotope of a single nucleus (stripped of its electrons) is accel

Question

The most common isotope of a single nucleus (stripped of its electrons) is accelerated from rest through a potential difference of 1000 V and then fired into a uniform 58 mT B-field (directed out of the page) such that the nucleus is traveling perpendicular to that B-field. The nucleus is observed to travel in a circular path of radius 11.4412 cm. There are several parts to this problem, but part c) What element is the unknown nucleus? - I can not figure out how to get the answer. I know how to find the velocity using the potential difference, and I know I need to use POCOE in some way, but I don't know the charge and I don't know the mass. I can't figure out a way to get either of those numbers, so without having a mass or a charge... how am I supposed to figure out what element it is?

Explanation / Answer

v^2 = (2qV/m)

m v^2/R = q v B

m v = R q B

m^2 v^2 = R^2 q^2 B^2

m^2 (2qV/m) = R^2 q^2 B^2

m (2qV) = R^2 q^2 B^2

m (2V) = R^2 q B^2

m = R^2 q B^2/(2 V)

m = q * 0.114412*0.114412 * 0.058*0.058/(2*1000)

==> m/q = 2.20176E-8 Kg/C

N = Number of nocleouses = mass/(proton mass)= mass/1.673e-27

Z = Number of protons = q/e = q/1.6e-19

==> m/q = (N * 1.673e-27)/(Z * 1.6e-19) = 2.20176e-8

==> N/Z = 2.1057

Number of nocleouses is two times the number of protons, terefore it is Helium. Because helium has 4 nocleouses and 2 protons.

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