0
$\begingroup$

Please take a look at (C). I have no prob with (a) and (b) but some values are needed from (a) and (b) in order to perform a calculation in (C).

For (C), I wonder why I got different answers from two different methods. I used binding energy per nucleon in method 1 and got 182 MeV in the blue box. However, with method 2, using the mass difference between products and reactants, I got 200 MeV instead (in the red box). After checking my answers with mark scheme, I found that the first one (182 MeV) is correct. Why method 2 is wrong?

Any help would be much appreciated.

Ps. I have checked my calculation again and again. Thus, I don't think it is about the numerical mistakes.

enter image description here

enter image description here

$\endgroup$
0
$\begingroup$

What is this strange textbook/test? Short answer: comparing binding energies -method 1- is wrong in this case. Use mass excess or amu.

First: I get quite another values, check it in literature...

B.E. per nucleon

139La 139,57 8.37806

95Mo 95,42 8.64867

235U 235,92 7.59091 MeV

Q reaction = 207.788 MeV (when I use mass excess differences -I like them more - or when I use a.m.u with 931.5 MeV as in your method 2)

Second:

The given question is not really just about a nuclear reaction. It is a reaction with number of subsequent decays. Seven $\beta-$ decays.

When I use your method of comparison of binding energies on both sides (method 1) I get 202.31014 MeV, which is also different from Q, similar problem as in your example.

What actually happens - I show on example of mass excess. Let us symbolically rewrite the reaction with masses (notation is mex(A,Z)):

$m_{ex}(235,92) \rightarrow m_{ex}(139,57) + m_{ex}(95,42) + m_{ex}(1,0)$

I omitted one neutron on each side for simplicity. If you substract the sides, you get 207.788 MeV.

Now we write the new formula with binding energies on each side and compare to the previous

$m_{ex}(235,92) - 92\cdot m_{ex}(1,1) - 143\cdot m_{ex}(1,0)\rightarrow m_{ex}(139,57) - 57\cdot m_{ex}(1,1) - 82\cdot m_{ex}(1,0)+ m_{ex}(95,42) - 42\cdot m_{ex}(1,1) - 53\cdot m_{ex}(1,0) + m_{ex}(1,0) - m_{ex}(1,0) $

just for illustration I left the last neutron binding energy explicitly. After a simplification you get

$m_{ex}(235,92) - 92\cdot m_{ex}(1,1) - 143\cdot m_{ex}(1,0)\rightarrow m_{ex}(139,57) + m_{ex}(95,42) - 99\cdot m_{ex}(1,1) - 135\cdot m_{ex}(1,0) $

If you wanted to get the same with mass excess a(or a.m.u.) AND binding energies, the terms with mex(1,0) and mex(1,1) must be the same on both sides. They are not.

Third:

If it was just a nuclear reaction ($^{235}U + n\rightarrow ^{139}I + ^{95}Y + 2n$), your binding energies formula would magically work in the case. In my case,I get the difference that is 5.477 MeV, which is exactly $7\times$ mass difference of proton and neutron $m_{ex}(1,0)-m_{ex}(1,1)$ =0.7823 MeV

$\endgroup$
  • $\begingroup$ Sorry for the very late reply. It's A-level exam. Thank you so much for your help. I am going to vote this answer up once I have enough reputations. Thank youuuu:))) $\endgroup$ – anonymous May 6 '17 at 2:45
0
$\begingroup$

Method 2 involved the condition counting the kinetic energy of neutrons and electrons which has been ignored by the question. That's the reason why you got two different answers.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.