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Physics - calculations

"The laser light is directed on to a small volume of deuterium fuel. This fuel is confined for a short time by its own inertia. The temperature increases considerably and fusion occurs for a brief instant converting deuterium into an isotope of helium.

2(2,1H) ---> (3,2He) + (1,0n)

The masses of the reacting nuclei are:

2H-----> Mass/u = 2.01410
3He-----> Mass/u = 3.01603
1n------> Mass/u = 1.00867

1 u = 1.66*10^-27 Kg.

Show that the number of atoms of deuterium involved in the fusion process is about 10^9 if the energy yield is only 10 millionths of the energy delivered by the laser (the efficiency is equal to 10^-5)"

I calculated the binding energy by using, DE = c^2Dm and the answer is 5.11*10^-13 J (answer is correct for that) Not sure what to do next here cause i don't really understand it tbh. If someone could explain to me what to do in the next bit please.

Thanks in advance.

OP

anyone?

OP

anyone now?

5.11*10^-13/input = 10^-5?

Reply 4

boromir9111

OP

17

Wesssty

5.11*10^-13/input = 10^-5?

Nope that don't work, i think. What makes you think it's that?

Reply 5

boromir9111

OP

17

that's the process is getting to me, cause now i have figured out the binding energy needed to turn the nucleons into a nucleus. As the post above mentioned we use output/input = efficiency. How would this able to give us the number of atoms involved? that's the part getting to me. Well we can't use that because they already given us the efficiency.

Reply 6

CHEM1STRY

11

boromir9111

that's the process is getting to me, cause now i have figured out the binding energy needed to turn the nucleons into a nucleus. As the post above mentioned we use output/input = efficiency. How would this able to give us the number of atoms involved? that's the part getting to me. Well we can't use that because they already given us the efficiency.

Could you work out the number of moles in each one and x it by Avogadro's constant?

Reply 7

boromir9111

OP

17

CHEM1STRY

Could you work out the number of moles in each one and x it by Avogadro's constant?

I would only need to do that with the deuterium wouldn't i?

edit - i don't think we can do that either cause in the mark scheme they do:

(40*10^-5/5.11*10^-13) *2....... i was hoping someone could explain where the other two numbers came from and why would they do that??

OP

anyone?

I don't know at what level this question is being set, but perhaps you're supposed to calculate the minimum energy required to bring the two deuterium together and use that as the energy of the laser?

Reply 10

boromir9111

OP

17

Fungrus

I don't know at what level this question is being set, but perhaps you're supposed to calculate the minimum energy required to bring the two deuterium together and use that as the energy of the laser?

This is at A2 level and i have calculated the energy, which is called the binding energy? or am i supposed to calculate something else as well? The whole point here is if someone has got the answer, please share the next step with me cause i have calculated that energy.

Reply 11

Stonebridge

13

boromir9111

I would only need to do that with the deuterium wouldn't i?

edit - i don't think we can do that either cause in the mark scheme they do:

(40*10^-5/5.11*10^-13) *2....... i was hoping someone could explain where the other two numbers came from and why would they do that??

Without knowing the power of the laser it can't be done.
The answer you quote above shows that they are using a value of 40J for the energy in the laser. (Multiply by efficiency to get actual power transferred, then divided by energy for one fusion to find how many fusions this could create.)
The original question starts "The laser light..." which suggests to me that this is a follow on from an earlier question about this laser. I can only assume that at some point earlier, either this value was given, or it was calculated.

Reply 12

boromir9111

OP

17

Stonebridge

Without knowing the power of the laser it can't be done.
The answer you quote above shows that they are using a value of 40J for the energy in the laser. (Multiply by efficiency to get actual power transferred, then divided by energy for one fusion to find how many fusions this could create.)
The original question starts "The laser light..." which suggests to me that this is a follow on from an earlier question about this laser. I can only assume that at some point earlier, either this value was given, or it was calculated.

Oh yeah, sorry about that, i calculated the power of the laser light earlier and it was 1*10^14W...... what do i do from here? thanks.

Reply 13

boromir9111

OP

17

So we have the power delivered by the laser and this is directed at the deuterium fuel and produces fusion, this provides the energy needed to separate the nucleons to form another nucleus. That energy calculated is 5.112*10^-13 J. We are given the efficiency of this energy.... but i don't get what to do next?

Reply 14

Stonebridge

13

boromir9111

Oh yeah, sorry about that, i calculated the power of the laser light earlier and it was 1*10^14W...... what do i do from here? thanks.

Is that value correct?
Without knowing the other part of the question about the laser we have no way of knowing.
Maybe you could post the whole question.
There is still some missing information.

Reply 15

boromir9111

OP

17

Stonebridge

Is that value correct?
Without knowing the other part of the question about the laser we have no way of knowing.
Maybe you could post the whole question.
There is still some missing information.

Yep, the value is correct, i was about to post it now.
" The possibility of nuclear fusion experiments in your living room is getting closer! Lasers could be used to deliver huge pulse of energy in a very short time to a small volume of deuterium (an isotope of hydrogen) to produce the conditions for fusion for just an instant.

A neodymium laser produces 40 J of coherent light of wavelength 1050nm for a period of 400 fs (1 fs = 10^-15s)"

Reply 16

boromir9111

OP

17

Is that sufficient enough info? if so, please walk me through it how to go about getting the final answer. Right now i have two marks out of the 4.... just not sure what to do next.

OP

anyone?

The efficiency is the proportion of that laser energy you can actually use.
It starts off as 40 J. The efficiency is 10^-5 (you can only use one 10 millionth of the 40 J, in other words.) So you multiply the 40 by 10^-5.
You calculated correctly that one fusion reaction needs 5.11 x 10^-13J
So how many fusion reactions will you get for your 40 x 10^-5 J ?

The *2 in the answer you gave earlier is there because the question asks "how many deuterium atoms". Each fusion reaction needs 2 deuterium.

Reply 19

boromir9111

OP

17

Stonebridge

The efficiency is the proportion of that laser energy you can actually use.
It starts off as 40 J. The efficiency is 10^-5 (you can only use one 10 millionth of the 40 J, in other words.) So you multiply the 40 by 10^-5.
You calculated correctly that one fusion reaction needs 5.11 x 10^-13J
So how many fusion reactions will you get for your 40 x 10^-5 J ?

The *2 in the answer you gave earlier is there because the question asks "how many deuterium atoms". Each fusion reaction needs 2 deuterium.

Quick question, why multiply by the 10^-5?

Physics - calculations

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