Alpha particle in gold foil scatering experiment

AI Thread Summary
The discussion centers on calculating how close an alpha particle approaches the nucleus during the gold foil scattering experiment, with an initial energy of 2.0 MeV. The user attempts to derive the distance using equations related to electric force and energy, ultimately calculating a distance of 1.13 x 10^-13 meters. However, they suspect an error since the expected answer should be around 4. A suggestion is made to divide by Joules instead of electron volts in the final calculation to correct the mistake. Accurate calculations are crucial for understanding the behavior of alpha particles in nuclear physics experiments.
asadpasat
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Homework Statement


How close does the alpha particle gets to the nucleus in gold foil scatering experiment?

Homework Equations


Ek = 2.0 MeV
W=F delta x
F=(kq1q2)/r^2
W=(kq1q2)/r
E=W=(kq1q2)/r

The Attempt at a Solution


E= (k(79)(2) e^2)/r
2.0MeV= (k(158)e^2)/r
2,000,000 eV = 8.99 Wm^2 c^-2 ((158) e^2) / r
2,000,000 eV convert by multiplying by 1.6 x 10^-19 J/ eV
r = 8.99 x10^ 9 Nm^2/c^2 (158) (1.6x 10^-19 c)^2 / 2,000,000 ev
r = 1.13 x 10^-13

I made somewhere mistake as the answer should be around 4 not 1.13 .
Any input appreciated
 
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asadpasat said:

Homework Statement


How close does the alpha particle gets to the nucleus in gold foil scatering experiment?

Homework Equations


Ek = 2.0 MeV
W=F delta x
F=(kq1q2)/r^2
W=(kq1q2)/r
E=W=(kq1q2)/r

The Attempt at a Solution


E= (k(79)(2) e^2)/r
2.0MeV= (k(158)e^2)/r
2,000,000 eV = 8.99 Wm^2 c^-2 ((158) e^2) / r
2,000,000 eV convert by multiplying by 1.6 x 10^-19 J/ eV
r = 8.99 x10^ 9 Nm^2/c^2 (158) (1.6x 10^-19 c)^2 / 2,000,000 ev
r = 1.13 x 10^-13
I made somewhere mistake as the answer should be around 4 not 1.13 .
Any input appreciated
How about dividing by Joules instead of eV in your last equation?
 
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