Nuclear reaction/binding energy problem? Year 11 physics, so its fairly simple.

[caitie_72]

Homework Statement

In the near future nuclear fusion reactions such as the one shown below may be used to produce electrical power.

2 4
2 H ----> He + energy
1 2

Physicists have suggested that 50 MW of energy will be sufficient to satisfy the energy needs of these industrial areas. If the nuclear reaction above is only 30% efficient, what mass of deuterium fuel will be needed per day.

atomic mass (21)H = 2.024202 amu and of (42)H = 4.002603

Homework Equations

The Attempt at a Solution

I'm in a state of panic right now. I have an exam tomorrow morning and don't get nuclear physics, it really just confuses me :(

Any help at all is appreciated. Please. Please. Please. I'm not one for begging but... please. (Please).

Thanks :)

My feeble attempt:

Find mass defect: H - H = 1.988501
* 1.6606*10^-27
= 2.1198 * 10^-27

E = mc^2
= 2.1198*10^-27 * (3*10^8)^2
= 1.9078 * 10^-10 j

Joules needed per day = 50*10^6 * 24 *60*60
= 4.32 *10^11

reactions per day =
4.32*10^11 / 1.9027*10^-10
2.26 * 10^21

total mass needed = 2.014 * 2.26*10^21
4.56 * 10^21 amu

I know I am probably wrong. I have no idea what I am doing. Its all guessing and confusion. I don't even have the answer to this problem, and its a revision sheet. Ugh. I am going to fail this exam so hard.

If you help me i will be eternally grateful. :)

 

[Andrew Mason]

atomic mass (21)H = 2.024202 amu and of (42)H = 4.002603

...

My feeble attempt:

Find mass defect: H - H = 1.988501
* 1.6606*10^-27
= 2.1198 * 10^-27

The mass defect would be the difference in mass of the reactant and the product nuclei. The reactants are 2 deuterium nuclei and the product is one He nucleus. So what is the change in mass (and, hence, the amount of energy released) per reaction?

Use that figure as the mass defect and the rest is fine.

AM

 

[thomate1]

Hello,

First of all, take a deep breath and try not to panic. Nuclear physics can be a challenging topic, but with some practice and understanding, you will be able to solve problems like these.

Let's start by breaking down the problem step by step. The first thing we need to do is to calculate the energy produced by the nuclear reaction. We can do this by using the formula E=mc^2, where m is the mass defect (the difference between the mass of the reactants and the mass of the products) and c is the speed of light.

From the given atomic masses, we can calculate the mass defect of the reaction as:
m = 4.002603 - 2.024202 = 1.978401 amu

Now, we need to convert this mass defect to kilograms in order to use the formula. We can do this by multiplying by the conversion factor 1.6606*10^-27 kg/amu:
m = 1.978401 * 1.6606*10^-27 kg/amu = 3.285 * 10^-27 kg

Plugging this value into the formula, we get:
E = (3.285 * 10^-27 kg) * (3*10^8 m/s)^2 = 2.9565 * 10^-10 J

This is the amount of energy produced by one reaction. Now, we can calculate the number of reactions needed to produce 50 MW of energy per day. Since 1 watt = 1 joule/second, 50 MW = 50*10^6 J/s. We can then calculate the number of reactions needed per second:
50*10^6 J/s / (2.9565 * 10^-10 J/reaction) = 1.69 * 10^17 reactions/s

To find the total number of reactions needed per day, we simply multiply by the number of seconds in a day (24*60*60):
1.69 * 10^17 reactions/s * (24*60*60 s/day) = 1.46 * 10^22 reactions/day

Finally, we can calculate the mass of deuterium fuel needed per day. Since each reaction requires 2 deuterium atoms, we can divide the total number of reactions by 2 to get the number of deuterium atoms needed:
1.46 * 10^