Initial energy of a charged particle

[skweeegor]

Homework Statement

Ernest Rutherford in his lab at McGill University, Montreal, fired alpha particles of mass 6.64 * 10^-27 kg at gold foil to investigate the nature of the atom. What initial energy must an alpha particle (charge +2e) have to come within 4.7 * 10^-15 m of a gold nucleus (charge +79e) before coming to rest? This distance is approximately the radius of the gold nucleus.

Givens:

m = 6.64 * 10^-27 kg

q_alpha = 2(1.6 * 10^-19 C)
= 3.2 * 10^-19 C

q_gold = 79(1.6 * 10^-19 C)
= 1.264 * 10^-17 C

r_gold = 4.7 * 10^-15 m

d = 4.7 * 10^-15 m

r = r_gold + d
= 9.4 * 10^-15 m

k = 9.0 * 10^9 N*m^2/C^2

Homework Equations

E = kq1q2/r

The Attempt at a Solution

So I know that the total energy of the system when the alpha particle is at rest and r distance from the centre of the gold nucleus is given by the equation E = kq1q2/r.

So, E = [(9.0*10^9)(3.2*10^-19)(1.264*10^-17)] / 9.4 * 10^-15

Which gives me E = 3.87 * 10^-12 J

However, this is exactly half of what the solution should be, and I can't seem to figure out why.

Thanks

 

[ehild]

Distance from the gold nucleus means the distance from its centre.

ehild

 

[skweeegor]

Yes, but the question asks what the initial energy of the particle would have to be when it is within 4.7 * 10^-15 m of a gold nucleus. It also states that 4.7 * 10^-15 is the radius of the gold nucleus. Therefore, the distance from the centre of the gold nucleus to the alpha particle is 2(4.7 * 10^-15 m). Am I misunderstanding the question?

 

[ehild]

Yes, but the question asks what the initial energy of the particle would have to be when it is within 4.7 * 10^-15 m of a gold nucleus. It also states that 4.7 * 10^-15 is the radius of the gold nucleus. Therefore, the distance from the centre of the gold nucleus to the alpha particle is 2(4.7 * 10^-15 m). Am I misunderstanding the question?

Usually the distance from a sphere-like object means the distance from its centre, especially for something which does has not a smooth surface.
When we speak about the distance of Moon from the Earth or the distance of Earth from the Sun, it is the distance between the centres, and not that between the surfaces. How would you take the atmosphere into account? Where does the Earth or the Sun end?
The gold nucleus contains a lot of protons and neutrons with no definite shape and not in rest. You can not compare the size of the nucleus by comparing it with a metre stick. The radius can be defined by an indirect way, buy scattering experiments fro example.

ehild

 

[asteriatic]

for your help!



Hello, thank you for sharing your work and question. Your approach and calculations are correct, so I believe the discrepancy in your answer may be due to rounding errors. I would suggest using the exact values for the charges and not rounding them off, as well as using the full value for the radius of the gold nucleus (4.7 * 10^-15 m) in your calculation. This should give you a more accurate result. Additionally, double check your units to make sure they are consistent throughout the calculation. I hope this helps. Keep up the good work in your studies!