Simple Potential Energy of an electric force prob

In summary, the conversation discusses the concept of nuclear fission and applies it to the uranium-236 nucleus. It explains that the nucleus can divide into two smaller nuclei with half the charge and volume, and this reaction was present in the nuclear weapon used in Hiroshima. The conversation then presents a model for the fission process and calculates the kinetic energy of the daughter nuclei when they are far apart. The solution involves using the equations for electric potential energy and kinetic energy, and dividing the total energy equally between the two nuclei.
  • #1
robbondo
90
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
[
 
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  • #2
SOLVED! divide by two!
 
  • #3
Response]

There are a few things to consider in your attempt to solve this problem. First, make sure you are using the correct units in your calculations. The charge of the uranium-236 nucleus should be written as +92e, not +46e. Also, the charge of the "daughter" nuclei should be half of that, so +46e. So, the correct equation for U1 should be:

U1 = (9*10^9) * [(92*1.6*10^-19)^2 / (5.90*10^-15 * 2)]

Also, it is important to note that the potential energy is a scalar quantity, meaning it only has a magnitude and no direction. So, you do not need to include the negative sign in your calculation. This may be why you are getting a different answer from the computer.

Finally, to find the kinetic energy of the "daughter" nuclei, you can use the conservation of energy equation:

K1 + U1 = K2 + U2

Since K1 and U2 are both zero, this simplifies to:

K2 = U1

Substituting in the correct value for U1, we get:

K2 = 4.15*10^-11 J

Therefore, each of the "daughter" nuclei will have a kinetic energy of 4.15*10^-11 J when they are very far apart.
 

Related to Simple Potential Energy of an electric force prob

1. What is Simple Potential Energy of an electric force prob?

The Simple Potential Energy of an electric force prob refers to the energy that is associated with an object due to its position in an electric field. It is a measure of the work required to move a charged object from one point to another in an electric field.

2. How is Simple Potential Energy of an electric force prob calculated?

To calculate the Simple Potential Energy of an electric force prob, the formula E = qV is used, where E is the potential energy, q is the charge of the object, and V is the potential difference between the initial and final points in the electric field.

3. What are some examples of objects with Simple Potential Energy in an electric field?

Some examples include charged particles such as protons and electrons, as well as charged objects like a balloon that has been rubbed against a fabric and becomes negatively charged.

4. How does the distance between charged objects affect Simple Potential Energy?

The Simple Potential Energy between two charged objects is directly proportional to the distance between them. As the distance increases, the potential energy decreases, and vice versa.

5. What is the relationship between Simple Potential Energy and Kinetic Energy?

Simple Potential Energy and Kinetic Energy are forms of energy that are interconvertible. When an object moves from a higher potential energy to a lower potential energy, its kinetic energy increases. Similarly, when an object moves from a lower potential energy to a higher potential energy, its kinetic energy decreases.

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