Question

A bare helium nucleus has two positive charges and a mass of 6.64 ✕ 10-27 kg.

(a) Calculate its kinetic energy in joules at 4.60% of the speed of light.
J
(b) What is this in electron volts?
eV
(c) What voltage would be needed to obtain this energy?
V

Answer 1

so for a) use \(KE=\dfrac{1}{2}m_{He}*v^2\)

b) 1 eV = \(1.6*10^{-19}J\)

c) i dont have my notes with the formulas, hold up

Answer 2

so far so good

Answer 3

@cro825 what do you get for the kinetic energy when you plug the numbers in ?

Answer 4

6.32e-13 J

Answer 5

then 3.94e6 eV for b

Answer 6

yes, that's what I have - so what about the last part ?

Answer 7

Would you use the formula for a parallel-plate capacitor for the last one?

so \(\Delta V=\dfrac{Fd}{q}\), @ProfBrainstorm ?

Answer 8

I don't think there's any need aaronq

Answer 9

hmm okay

Answer 10

v=pe/q?

Answer 11

if a charge q drops thru a potential v then it acquires kinetic energy equal to qv

Answer 12

so you would need a voltage v = kinetic energy/2, if ke is in eV since there's a double charge on the helium nucleus

Answer 13

so i think you have it right cro825

Answer 14

voltage needed would be about 1.97 Megavolts

Answer 15

so ke/ev?

Answer 16

if the kinetic energy is measured in electron volts then charge is measured in units of the electron charge

Answer 17

no 3.94e6/2

Answer 18

got it thanks guys

Answer 19

ok

Answer 20

np!

Answer 21

In case you're interested, the formula for kinetic energy mv^2/2 only works when v is a small fraction of c, the speed of light, (which is fine for this question).
But if the helium nucleus was moving significantly faster you would have to use the correct expression from special relativity for the energy

Answer 22

ok cool