Please provide step by step instructions and print the response so I can read the response.
1) Consider the reaction
a) Use the masses of the nuclides involved to determine where it
is endogenic of exoergic.
b) If it is exoergic, find the amount of energy released, if it is
exoergic find the threshold energy.
NOTE THAT IN THESE TYPE OF PROBLEMS TRY TO WRITE AS MANY DECIMAL PLACES AS POSSIBLE TO GET ACCURATE VALUES.
Mass of Nucleus, C = 13.0033548377 u
Mass of H = 1.00782503207 u
Mass of He Nucleus = 4.00260325415 u
Mass of Be = 10.01293699 u
Therefore,
energy released is
E = 13.0033548377 + 1.00782503207 - (4.00260325415 + 10.01293699)
E = 14.01117987 - 14.01554024
E = - 0.00436037415 * 931.5
As we can see that E is negative, so, reaction is endogenic.
We will find its threshold energy
E (threshold) = (total mass * mass defect / 2 * nucleus mass (C))*931.5 MeV
E (threshold) = 4.699056e-3 * 931.5
E (threshold) = 4.377 MeV
Please provide step by step instructions and print the response so I can read the response....
Please provide step by step instructions and print so I can read the response. 1) Find the threshold energy for the following reaction: 2) Calculate the energy released in the fission reaction:
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HELLO I AM CURRENTLY IN DIFFERENTIAL EQUATIONS PLEASE EXPLAIN EACH STEP SO I CAN LEARN FROM YOU (I KNOW SOME PEOPLE ONLY CARE ABOUT TEH ANSWER, BUT WILL REALLY APPRECIATE IT) TO SAVE TIME FEEL FREE TO JUST SAY A LAW, THEOREM, OR CONCEPT FOR AN EXPLANATION AND I CAN GO AHEAD AND STUDY IT ON MY OWN. i REALLY DO READ THESE VERY CAREFULLY AND USE THE COMMENTS OFTEN, SO JUST A LITTLE HEADS UP. I FIND IT DIFFICULT...
HELLO I AM CURRENTLY IN DIFFERENTIAL EQUATIONS PLEASE EXPLAIN EACH STEP SO I CAN LEARN FROM YOU (I KNOW SOME PEOPLE ONLY CARE ABOUT TEH ANSWER, BUT WILL REALLY APPRECIATE IT) TO SAVE TIME FEEL FREE TO JUST SAY A LAW, THEOREM, OR CONCEPT FOR AN EXPLANATION AND I CAN GO AHEAD AND STUDY IT ON MY OWN. i REALLY DO READ THESE VERY CAREFULLY AND USE THE COMMENTS OFTEN, SO JUST A LITTLE HEADS UP. I FIND IT DIFFICULT...
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