The attraction force provides the electron's centripetal force.
<span>8.30^-8N = mrω² </span>
<span>ω² = 8.30^-8 / (9.11^-31 kg x 4.70^-11m) .. .. ω² = 1.94^33 (rad/s)² .. .. ω = 4.40^16 rad/s </span>
<span>f = ω/2π = 4.40^16 / 2π .. .. .. ►f = 7.0^15 Hz</span>
Answer: Ok, first lest see out problem.
It says it's a Long cylindrical charge distribution, So you can ignore the border effects on the ends of the cylinder.
Also by the gauss law we know that E¨*2*pi*r*L = Q/ε0
where Q is the total charge inside our gaussian surface, that will be a cylinder of radius r and heaight L.
So Q= rho*volume= pi*r*r*L*rho
so replacing : E = (1/2)*r*rho/ε0
you may ask, ¿why dont use R on the solution?
since you are calculating the field inside the cylinder, and the charge density is uniform inside of it, you don't see the charge that is outside, and in your calculation actuali doesn't matter how much charge is outside your gaussian surface, so R does not have an effect on the calculation.
R would matter if in the problem they give you the total charge of the cylinder, so when you only have the charge of a smaller r radius cylinder, you will have a relation between r and R that describes how much charge density you are enclosing.
If its atomic number is 48, then it has 48 protons in the nucleus
of each atom. Any more mass than that is supplied by the neutrons
that are mixed in there with the protons.
If the mass is 167, and 48 of those are protons, then there are
(167 - 48) = 119 neutrons
in each nucleus.
