Calculating Energy Released from Uranium Fission

[nima rahmani]

hey guys i have this question which says:
On average the fission of a uranium 235 nucleus releases 200 MeV of energy. How many

joules of energy will be released by the fission all the uranium 235 atoms in our fuel.

my answer: (200*10^6)*1.6*10^-19 which is 3.2*10^-13.

but when i compered with my friend's works, they were all different. one of them was 3.2*10^-20
and the interesting thing is that when i googled it it said: 3.2*10^-11.

i am confused now and need someone to help me out with this.
thanks

 

[MarcusAgrippa]

This is a standard problem in unit conversions. The best and most foolproof method to use is called the chain rule. It works like this. Write your units next to the number as they were a fraction. For example,
150 km/hr = 150 $\frac{km}{hr}$
Next, write the simplest relation between the units you want to convert. Suppose 1 want to convert km/hr to m/s. First, the relation between m and km is
$1 {\rm km} = 10^3 {\rm m}$
To get from hr to s, we have two easy relations
$1 {\rm hr} = 60 {\rm min} \ \mbox{ and }\ 1 {\rm min} = 60 {\rm s}$
Each of these relations can be written in 2 ways as a fraction. For example,
$1 {\rm km} = 10^3 {\rm m} \ \Leftrightarrow\ \frac{1 {\rm km}}{10^3 {\rm m}} = 1 \ \Leftrightarrow\ \frac{10^3 {\rm m}}{1 {\rm km}} = 1$

Now for the conversion: string together these fractions with your starting quantity in such a way as to cancel the unwanted units and leave the units that are wanted. Each fraction is equal to 1, so you have not changed the original quantity. The above example then becomes
150 km/hr = 150 $\frac{km}{hr} \times \frac{10^3 {\rm m}}{1 {\rm km}} = 150 \times 10^3 \frac{\rm m}{\rm hr}$
In the original quantity, km appears in the numerator. To get rid of it in favour of m, I need to use the conversion fraction with km in the denominator, to cancel the m in the numerator. This leaves m in the numerator. We now get rid of the hr in favour of s,
150 km/hr = 150 $\frac{km}{hr} \times \frac{10^3 {\rm m}}{1 {\rm km}} \times \frac{\rm 1 hr}{60 {\rm min}} \times \frac{1 {\rm min}}{60 {\rm s}} =$
Cancelling the units and collecting the numbers we get
150 km/hr = $150 \times \frac{10^3}{60 \times 60} \frac{\rm m}{\rm s}$
The units are now correct. all you need do is tidy up the numbers.

Try this method for your example. It will give you the correct answer, if you apply the method correctly, and there can be no controversy about the answer. The relations that you need are
$1 {\rm MeV} = 10^6 {\rm eV}\ \mbox{ and }\ 1 {\rm eV} = 1.609 \times 10^{-19} {\rm J}$

 

[SteamKing]

hey guys i have this question which says:
On average the fission of a uranium 235 nucleus releases 200 MeV of energy. How many

joules of energy will be released by the fission all the uranium 235 atoms in our fuel.

my answer: (200*10^6)*1.6*10^-19 which is 3.2*10^-13.

You need to check your arithmetic in this previous calculation.

but when i compered with my friend's works, they were all different. one of them was 3.2*10^-20
and the interesting thing is that when i googled it it said: 3.2*10^-11.

i am confused now and need someone to help me out with this.
thanks

One of the reasons your work is confusing you (and confusing to others) is you do not show the units of the quantities in your calculations.

You should get into the habit now of showing units so that your work is clear.

Also, go over your calculations and make sure you have not committed any silly arithmetic errors as you have done here.
It will save you time and worry in the long run, and it may keep you from losing points on exams and other work done for credit.

 

[MarcusAgrippa]

Remind me not to waste time answering questions from students who are not serious about their work. The "hey guys" opening should have alerted us to the fact that we were wasting our time.