Calculating energy released in nuclear fissionConsider the neutron induced fission U-235 + n → ⋯ → La-139 + Mo-95 + 2 n, where … denotes intermediate decay steps.I want to calculate the released energy from this fission. One way would be to calculate the difference of the binding-energies ( B): Δ E = B ( 139 , 57 ) + B ( 95 , 42 ) − B ( 235 , 92 ) ≈ 202 , 3 M e V (Btw. I didn't use the binding energies from a semi-empirical binding energy formula but calculated them directly via mass defect).Another way is: Δ E = ( m ( U-235 ) + m Neutron − m ( Mo-95 ) − m ( La-139 ) − 2 m Neutron ) c 2 ≈ 211 , 3 M e V Which one gives the correct result? Why?You notice that 57 + 42 ≠ 92, if that was the case, it would be equal, but I don't clearly see where the difference physically comes from and what to add or subtract (and why) from to the first or from the second term to get the other result. How to make this clear?A slightly other point of view: What different questions do both calculations answer?

Question

Calculating energy released in nuclear fissionConsider the neutron induced fission   U-235  +  n  →  ⋯  →  La-139  +  Mo-95  +  2  n, where   … denotes intermediate decay steps.I want to calculate the released energy from this fission. One way would be to calculate the difference of the binding-energies (  B):  Δ  E  =  B  (  139  ,  57  )  +  B  (  95  ,  42  )  −  B  (  235  ,  92  )  ≈  202  ,  3        M    e    V  (Btw. I didn&#039;t use the binding energies from a semi-empirical binding energy formula but calculated them directly via mass defect).Another way is:  Δ  E  =  (  m  (  U-235  )  +      m          Neutron        −  m  (  Mo-95  )  −  m  (  La-139  )  −  2      m          Neutron        )      c    2    ≈  211  ,  3        M    e    V  Which one gives the correct result? Why?You notice that   57  +  42  ≠  92, if that was the case, it would be equal, but I don&#039;t clearly see where the difference physically comes from and what to add or subtract (and why) from to the first or from the second term to get the other result. How to make this clear?A slightly other point of view: What different questions do both calculations answer?

Giancarlo Shah · Accepted Answer

Below I show how the discrepancy   (  202.3        M    e    V    ) and   (  211.3        M    e    V    ) between your two methods has arisen.  B  (  139  ,  57  )  =      57    p    +    82    n    +    57    e    −  m  (  139  ,  57  )  B  (  95  ,  42  )  =      42    p    +    53    n    +    42    e    −  m  (  95  ,  42  )  B  (  235  ,  92  )  =      92    p    +    143    n    +    92    e    −  m  (  235  ,  92  )  B  (  139  ,  57  )  +  B  (  95  ,  42  )  −  B  (  235  ,  92  )  =  m  (  235  ,  92  )  −  m  (  139  ,  57  )  −  n  +  7  e  +  [  7  p  −  7  n  ]  7  p  −  7  n  =  7  (  938.272  −  939.565  )  =  −  9.051There is your difference.Note that your original equation was unbalanced if you used the masses of the atoms because there is a difference of seven electrons between the left hand side   (  U  235  +  n  ) and the right hand side   (  L  a  139  +  M  o  95  +  2  n  ) of your equation.

dresu9dnjn · Accepted Answer

The released energy is the difference in energy of intitial/end products. For a nucleus, the energy is given by its mass, which in turn can be calculated as the difference of &quot;naive mass&quot;, i.e. all single constituents&#039; masses summed up, and the binding energy. This is the calculation you&#039;ll have to do. Obviously, all these calculations neglect kinetic energy terms.

Calculating energy released in nuclear fission Consider the neutron induced fission <mtext>U-2

Answered question

Answer & Explanation

New Questions in Nuclear physics