Simple Potential Energy of an electric force prob

Click For Summary
SUMMARY

The discussion focuses on calculating the kinetic energy of daughter nuclei resulting from the fission of uranium-236, modeled as a uniformly charged sphere with a charge of +92e and a radius of 7.4 x 10-15 m. The correct approach involves using the potential energy formula U = (1/4πε0) * (q * q0) / r, where the charges of the daughter nuclei are +46e each. The user initially calculated the potential energy incorrectly as 4.15 x 10-11 J but later realized the mistake was in not dividing by two, leading to the correct kinetic energy calculation.

PREREQUISITES
  • Understanding of nuclear fission processes
  • Familiarity with electrostatic potential energy equations
  • Knowledge of kinetic energy calculations
  • Basic grasp of charge and radius concepts in physics
NEXT STEPS
  • Study the principles of nuclear fission and its applications in energy generation
  • Learn about electrostatic potential energy in charged spheres
  • Explore the relationship between potential energy and kinetic energy in physics
  • Investigate the historical context and implications of the Hiroshima nuclear explosion
USEFUL FOR

Students studying nuclear physics, educators teaching concepts of energy transformations, and anyone interested in the mechanics of nuclear fission and its historical significance.

robbondo
Messages
90
Reaction score
0

Homework Statement


The unstable nucleus uranium-236 can be regarded as a uniformly charged sphere of charge Q = + 92e and radius [tex]R = 7.4 \times 10^{ - 15} {\rm m}[/tex]. In nuclear fission, this can divide into two smaller nuclei, each of [tex]{\frac{1}{2}}[/tex] the charge and [tex]{\frac{1}{2}}[/tex] the volume of the original uranium-236 nucleus. This is one of the reactions that occurred in the nuclear weapon that exploded over Hiroshima, Japan in August 1945. In a simple model for the fission process, immediately after the uranium-236 nucleus has undergone fission the "daughter" nuclei are at rest and just touching. Calculate the kinetic energy that each of the "daughter" nuclei will have when they are very far apart.


Homework Equations



[tex]U = \frac{1}{4\pi\epsilon_{0}}\frac{q * q_{0}}{r}[/tex]

[tex]K = 1/2 m v^{2}[/tex]

[tex]U_{1} + K_{1} = K_{2} + U_{2}[/tex]

The Attempt at a Solution



so I know for a fact because I got the answer correct(online hw) that the radius of the new spheres is 5.90*10^-15 m.
I know that K1 is zero because they are at rest, and I know that U2 is zero because they are "far" apart. So I solve for U1=K2

so for U1 I multiply the charges of +46E as stated in the problem divided by the radius time two, the distance between and then multiply that by the constant value 9*10^9

[tex]9*10^{9} \frac{(46*1.6*10^{-19})^{2}}{5.90*10^{-15}*2}[/tex]

So I get 4.15*10^-11 J which is WRONG according to this computer... what the heck am I doin' wrong
[
 
Physics news on Phys.org
SOLVED! divide by two!
 

Similar threads

  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
6
Views
2K
Replies
15
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 43 ·
2
Replies
43
Views
4K
  • · Replies 9 ·
Replies
9
Views
5K