Alpha particle in gold foil scatering experiment

Click For Summary
SUMMARY

The discussion focuses on calculating the distance an alpha particle approaches the nucleus during the gold foil scattering experiment. The initial calculations used the kinetic energy of 2.0 MeV and the formula for electric potential energy, leading to an incorrect result of 1.13 x 10^-13 meters. Participants suggest revisiting the conversion from electron volts (eV) to joules (J) to correct the calculation, as the expected distance should be around 4 x 10^-14 meters.

PREREQUISITES
  • Understanding of kinetic energy in particle physics
  • Familiarity with Coulomb's law and electric force calculations
  • Knowledge of energy conversion between electron volts and joules
  • Basic grasp of the gold foil experiment and its significance in atomic structure
NEXT STEPS
  • Review the conversion process from electron volts to joules in energy calculations
  • Study Coulomb's law and its application in particle interactions
  • Learn about the significance of the gold foil experiment in understanding atomic structure
  • Explore advanced topics in nuclear physics related to alpha particle behavior
USEFUL FOR

Students studying nuclear physics, educators teaching particle interactions, and anyone interested in the principles of atomic structure and scattering experiments.

asadpasat
Messages
41
Reaction score
1

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
 
Physics news on Phys.org
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?
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 17 ·
Replies
17
Views
15K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
17K
  • · Replies 3 ·
Replies
3
Views
15K