Back
to top
Nuclear Physics Exam Question Help Needed
Watch
Announcements
Page 1 of 1
Go to first unread
Skip to page:
Just came across a 1 mark question on Paper 2 June 2018 and I am struggling to find the answer. Could you provide some hints that will help me reach the answer please?
Question :
During a single fission event of uranium-235 in a nuclear reactor the total mass lost is 0.23 u. The reactor is 25% efficient.
How many events per second are required to generate 900 MW of power?
Question :
During a single fission event of uranium-235 in a nuclear reactor the total mass lost is 0.23 u. The reactor is 25% efficient.
How many events per second are required to generate 900 MW of power?
0
reply
Report
#2
(Original post by I-ZAAA)
Just came across a 1 mark question on Paper 2 June 2018 and I am struggling to find the answer. Could you provide some hints that will help me reach the answer please?
Question :
During a single fission event of uranium-235 in a nuclear reactor the total mass lost is 0.23 u. The reactor is 25% efficient.
How many events per second are required to generate 900 MW of power?
Just came across a 1 mark question on Paper 2 June 2018 and I am struggling to find the answer. Could you provide some hints that will help me reach the answer please?
Question :
During a single fission event of uranium-235 in a nuclear reactor the total mass lost is 0.23 u. The reactor is 25% efficient.
How many events per second are required to generate 900 MW of power?
Then divide 900MW by that number and you get the number of ‘events’ per second (I think?)
Hope this helps : ))
1
reply
(Original post by akay8)
My first thought is mass energy equivalence - using E=mc^2 to work out the energy of 0.23u then multiply that by 0.25 as the reactor is only 25% efficient.
Then divide 900MW by that number and you get the number of ‘events’ per second (I think?)
Hope this helps : ))
My first thought is mass energy equivalence - using E=mc^2 to work out the energy of 0.23u then multiply that by 0.25 as the reactor is only 25% efficient.
Then divide 900MW by that number and you get the number of ‘events’ per second (I think?)
Hope this helps : ))
0
reply
Report
#4
(Original post by I-ZAAA)
Thank You that really helped but can you tell me why did you divided 900MW by that number please?
Thank You that really helped but can you tell me why did you divided 900MW by that number please?
Power is in watts which are J s^-1. Your energy calculated from E=mc^2 x 0.25 is measured in joules (J). Do the division (J s^-1)/J, the Js cancel, and you get s^-1, which is the number of events ‘per second’ ie what the question is asking for.
Basically you have a total energy and the energy of 1 nucleus, so doing total/(1 nucleus) = number of nuclei : ))
Last edited by akay8; 1 year ago
0
reply
(Original post by akay8)
I think about physics in terms of units
Power is in watts which are J s^-1. Your energy calculated from E=mc^2 x 0.25 is measured in joules (J). Do the division (J s^-1)/J, the Js cancel, and you get s^-1, which is the number of events ‘per second’ ie what the question is asking for.
Basically you have a total energy and the energy of 1 nucleus, so doing total/(1 nucleus) = number of nuclei : ))
I think about physics in terms of units
Power is in watts which are J s^-1. Your energy calculated from E=mc^2 x 0.25 is measured in joules (J). Do the division (J s^-1)/J, the Js cancel, and you get s^-1, which is the number of events ‘per second’ ie what the question is asking for.
Basically you have a total energy and the energy of 1 nucleus, so doing total/(1 nucleus) = number of nuclei : ))
0
reply
Report
#6
(Original post by I-ZAAA)
oooooooooh makes sense. THANK YOU
oooooooooh makes sense. THANK YOU
Silly question. Is that the right answer lol? Do you have the mark scheme?
0
reply
(Original post by akay8)
Haha np.
Silly question. Is that the right answer lol? Do you have the mark scheme?
Haha np.
Silly question. Is that the right answer lol? Do you have the mark scheme?
0
reply
X
Page 1 of 1
Go to first unread
Skip to page:
Quick Reply
