Electric Potential Problem Involving Uranium Atom

[Kumo]

Hi Everybody,

I am currently battling with the fourth and final question of my A Level physics work for this week. The problem has been driving me crazy, as I seemed to have completed all of the relevant steps, and I find the right answer only with negative exponents, which isn’t much help at all.

Homework Statement

In a simplified model, a uranium nucleus is a sphere of radius 8.0 * 10^-15m. The nucleus contains 92 protons (and rather more neutrons). The charge on a proton is 1.6 * 10^-19. It can be assumed that the charge of these protons acts as if it were all concentrated at the centre of the nucleus. The nucleus releases an α-particle containing two protons (and two neutrons) at the surface of the nucleus.

It then proceeds to ask questions for the electric force on the α-particles at the surface, and the electric potential of the nucleus, which I was fine with after a few tries. However, question (d) asks the following.

Calculate: the electric potential energy of the α-particle when it is at the surface of the nucleus.

Homework Equations

$W=\frac{kqQ}{r}$ or $E= \frac{V}{d}$ together with $V=\frac {kQ}{r}$

The Attempt at a Solution

I decided to tackle the problem in two ways, the first is as follows using the electric potential energy formula given above.

$W = \frac{qkQ}{r}$

$W = \frac{(8.99 * 10^9) * (3.2 *10^{-19}) * (1.44 * 10^{-17})}{(8 * 10^{15})}$

$W = \frac{(4.143 * 10^{-26})}{(8 * 10^{15})}$

$W=5.18 * 10^{-12}$ J

This is pretty much the same answer that I found in the back of the book, only with a negative exponent.

My next attempt was a bit of a shot in the dark, but I tried it regardless.

$V = \frac{kQ}{r}$

$V = \frac{(8.99 * 10^9) * (3.2 * 10^{19})}{(8.0 * 10^{19})}$

$V = \frac{(2.877 * 10^{-19}}{(8 * 10^{-19}}$

$V = \frac{3.6 * 10^{-1}}V$

$E = \frac{V}{d}$

$E = \frac{(3.6 * 10^{-1})}{(7*10^{-19}}$

$E = 4.5 * 10^{18} J$ J

I would really appreciate it if someone could shed some light on this issue, as it has really been bugging me. Thank you very much for any help and your time.

 

[mfb]

There are sign errors in your exponents.

How do you get 8.0*10¹⁹ in the second approach?
Some other exponents there look odd, too (how did you combine 9 and 19 to get -19?), and your units are missing or wrong.

Just be more careful with your numbers and units, and all those mysteries will disappear.

 

[Kumo]

Thanks for the reply, mfb. The second solution was copied from the document that I was using for my work, and I had attempted that method numerous times, so I must have made a mistake with that approach. A fixed version can be seen below. The second solution seems to be completely wrong. I apologize for posting such incorrect workings earlier.


$V = \frac{kQ}{r}$

$V = \frac{(8.99 * 10^9) * (3.2 * 10^{-19})}{(8.0 * 10^{-15})}$

$V = \frac{(2.877 * 10^{-9}}{(8 * 10^{-15}}$

$V = {3.6 * 10^{5}}$ V

$E = \frac{V}{d}$

$E = \frac{(3.6 * 10^{5})}{(8*10^{-15}}$

$E = 4.5 * 10^{19}$

 

[mfb]

First, the formula E=V/d assumes that E is constant, and V is zero at a distance of d. Both are wrong.
Second, why do you try to calculate the electric field here?
Third, please add units to your calculations.

 

[Kumo]

In hindsight I shouldn't have included the second attempt , as even in my own mind it didn't seem right. What would the main issue be with the main attempt at a solution? Am I using the incorrect formula?

 

[mfb]

Apart from the mentioned errors in the exponents, it looks right.
Why do you think it would be wrong?

 

[Kumo]

It's strange because the back of the book has the answer as 5.18*10¹². So I was able to get the right answer, only with a negative exponent.

 

[mfb]

Probably a typo in the book.
5.18*10¹² J is roughly the energy you get in the explosion of 1000kg of TNT - certainly not the energy released in a single nuclear reaction!

 

[Kumo]

Probably a typo in the book.
5.18*10¹² J is roughly the energy you get in the explosion of 1000kg of TNT - certainly not the energy released in a single nuclear reaction!

I thought that it seemed rather odd, but I thought that I had to be overlooking something. I did think that it was an awfully large number given the problem. Thank you very much for your help.