Nulcear fission, two daughter nuclei

[Taylor_1989]

Homework Statement

An unstable nucleus with mass M is in an excited state with excitation energy E* and undergoes spontaneous fission into two daughter nuclei "1" and "2" and zero fast neutrons. What will be the kinetic energy KE (in MeV) of the daughter nucleus "1"?

M = 233.9493amu
E* = 10.6616Mev
M1 = 123.3604 amu

M2 = 110.5096 amu

Homework Equations

Conservation of momentum
Total kinetic energy

The Attempt at a Solution

Here is my solution but I believe I am on the right line IMO but I not entirely sure what the activation energy really is and also why M = 233.9493amu has been included as it not in my solution which indicated to me I have gone maybe slightly wrong.

$$M_1V_1=M_2V_2 [1]$$
$$E^*=1/2M_1V_1^2+1/2M_2V_2^2 [2]$$
$$\frac{M_1V_1}{M_2}=V_2\:\left[3\right]$$
$$E^*=\frac{1}{2}M_1V_1^2+\frac{1}{2}\frac{M_1^2V_1^2}{M_2^2}\cdot M_2 [4]$$

Simplifying through and making $V_1^2$ the subject

$$\frac{2E^*}{\left(M_1+\frac{M_1^2}{M_2}\right)}=V_1^2$$

Subbing in the given values
$$\frac{2\left(10.6616MeV\right)}{\left(123.3604amu+\frac{\left(123.3604amu\right)^2}{110.5096amu}\right)}=V_1^2=0.08167\frac{Mev}{amu}\:\left[5\right]$$

$$V_1^2=0.08167\frac{Mev}{\frac{931.5MeV}{c^2}}=0.000087c^2 [6] $$

$$KE_1=\frac{1}{2}\left(123.3604amu\cdot 0.000087c^2\right)=\frac{1}{2}\left(123.3604\cdot 931.5\frac{MeV}{c^2}\cdot 0.000087c^2\right) [7]$$

this gave me a $KE$ value of $4.998MeV$

this issue is I don't really understand the excitation energy with regards to fission, so I a assume this energy is used to intaite the fission process.

 

[drvrm]

M = 233.9493amu
E* = 10.6616Mev
M1 = 123.3604 amu

M2 = 110.5096 amu

looking at your data it appears that there is a mass difference in mother and daughters nuclei added together ...so energy conservation must account for this part...

most of the mass difference goes into K E of the fragments.

 

[Taylor_1989]

looking at your data it appears that there is a mass difference in mother and daughters nuclei added together ...so energy conservation must account for this part...

most of the mass difference goes into K E of the fragments.

I what you are saying to add the mass diff to the excitation energy, accuse I did have this thought but the difference is a negative value

 

[drvrm]

difference is change in the masses and equivalent energy must be accounted for in conservation of energy...check from your textbook

.if the daughters are playing..the mother must have provided energy. just talking lighty.

 

[haruspex]

I a assume this energy is used to intaite the fission process.

Yes, but it is not consumed by the fission process. It still ends up in the KE.

add the mass diff to the excitation energy, accuse I did have this thought but the difference is a negative value

Which way round did you take the difference? Please post your working.

 

[Taylor_1989]

Yes, but it is not consumed by the fission process. It still ends up in the KE.

Which way round did you take the difference? Please post your working.

Sorry for the confusion when I looked over the question again I took the difference the wrong way round, so my orginall working was
$$(M_1+M_2)-M=\Delta M$$ but what I should have done is $$M-(M_1+M_2)=\Delta M$$. Which after this I got the correct ans.