1.Write
out the full nuclear reaction for:

_{13 }

*Al*^{27}(

*a*

*, n)*_{15}P^{30}
Thence,
or otherwise, find the Q of the reaction.

__Solution__: We write out the reaction from its condensed form given:

_{13 }

*Al*^{27}+_{2}He^{4 }**®**

_{15}P^{30 }+_{0 }n^{1}
Then:
The Q of the reaction is defined according to:

Q
= [M

_{a }+ M_{X}- M_{Y}– M_{b}]c^{2}
Where:

_{0 }**M**

*n*^{1}=_{b }

*= 1.008665u*

*a***M**

*=*_{X }

*=*_{2}He^{4 }= 4.002603u

_{15}

*P*^{30 }**M**

*=*_{Y}

*=*

*29.9783138u (from Wikipedia table, Googled)*

_{13 }**M**

*Al*^{27}=_{a }

*=***26.9815386u (from Wikipedia table – Googled)**

**Q =**[26.9815386u + 4.002603u - 29.978313 u - 1.008665u]

**931 MeV/u**

**=**[30.984141 u - 30.986978]u

**(**931 MeV/u)= - 2.64 MeV

(Negative
sign denotes an endothermic reaction)

2.
Identify the missing ‘X’ in each of the following:

a)

Po _{84 }^{215}*®**X +**a*
b)

_{ }N^{14}(*a**, X) O*^{17}
c)

_{48}Cd^{109}+ X*®*_{47 }Ag^{109}+*u*__Solutions__:

a)
This
is an example of

*alpha decay*defined by:

_{Z }

*X*^{A}

*®*

_{Z-2}X^{A-4}+_{2}He^{4}
Then: A – 4 = 215 – 4 = 211

And: Z – 2 =
84 – 2 = 82

So:
X = Pb (lead) e.g.

_{82 }Pb^{211}
b)
This is a condensed form for the reaction which would be written out:

_{7 }

*N*^{14}+

*a*

*®*

*X +*_{8 }O^{17}
Or:

_{7 }N^{14}*+*_{2}He^{4 }*®*_{Z1 +Z2}X^{ A1 + A2}+_{8 }O^{17}
Now: A1 + A2 = 14 + 4 = 18 and Z1 + Z2 = 7 +
2 = 9

So:

**which is**_{Z1 +Z2}X^{ A1 + A2}=_{9 }X^{18 }**fluorine**, or X = F (e.g.**)**_{9 }F^{18 }
b)
By
inspection we see the mass number Z remains unchanged but the atomic number
decreases by one. This must be an example of beta decay, defined:

_{Z}

*X*^{A}+_{-1}e^{0}

*®*

_{Z-1}X^{A}+

*u*
Hence,
X =

_{-1}e^{0}
So
that:

_{48}Cd^{109}+_{-1}e^{0}*®*_{47 }Ag^{109}+*u*
3.
Consider a process of neutron removal whereby:

_{20}

*Ca*^{43}

*®*

_{20 }Ca^{42}+_{0 }n^{1}
We
wish to find the “

*separation energy*”**associated with the neutron. Estimate the value of this energy if the atomic mass of***S*_{n}**= 42.98780u, and***Ca*^{43}**= 41.958625u, and take the neutron mass = 1.008665u.***Ca*^{42}**is just another variant of the Q for a reaction where:**

*S*_{n}**=**

*S*_{n}

*[(total rest mass before decay) –*

*(total rest mass after decay )] c*^{2}**=**

*S*_{n}[(42.98780 u) - (41.958625u + 1.008665u)] 931 Mev/ u =

(42.98780
u – 42.96729u ) 931 Mev/ u =

0.0205 u (931 Mev/ u ) = 19.09 MeV

0.0205 u (931 Mev/ u ) = 19.09 MeV

4.
For each of the following reactions, write out the full nuclear equation and
find the Q of the reaction:

a)

**(Note: d is for***H*^{2}(d, p) H^{3 }*deuterium*or_{1 }H^{2})
b)

*Li*^{7}(p, n) Be^{7}__Solution:__

a)

**+**_{1}H^{2 }_{1}H^{2 }*®***+**_{1}H^{1 }_{1}H^{3 }
Q
= [M

_{a }+ M_{X}- M_{Y}– M_{b}]c^{2}

_{1 }**M**

*H*^{2}=_{a }

*= 2.01410 u*

_{1 }**M**

*H*^{2}=_{X }

*= 2.01410 u*

_{1}**M**

*H*^{1 }=_{b }

*= 1.007825 u*

^{ }**M**

_{1 }H^{2}=_{Y}

*= 3.01604 u (from Wikipedia)*
Therefore:

Q
= [2(2.01410
u) - 1.007825 u - 3.01604 u] 931 Mev/u

= [(4.0282) – 4.0238)] 931 Mev/u =
0.0044 (931 Mev/u ) = 4.09 MeV

5.
Refer to the example of the fission of U 235 illustrated at the outset of Part
1. This fission reaction may be written:

_{92}

*U*^{235}+_{0 }n^{1}

*®*

_{42}Mo^{95}+_{57}La^{139}+ 2 (_{0 }n^{1})
Find
the Q-value of this reaction, given these atomic masses:

^{ }

_{0 }n^{1}= 1.009u

_{92}

*U*^{235}= 235.123u

_{42}

*Mo*^{95}= 94.945u

_{57}

*La*^{139}= 138.955u
Q
= [M

_{a }+ M_{X}- M_{Y}– M_{b}]c^{2}
Where:

_{92}**M**

*U*^{235}=_{a }

*= 235.123u*

_{0 }**M**

*n*^{1}=_{X }

*= 1.009u*

_{42}**M**

*Mo*^{95}+_{57}La^{139}^{ }=_{b }

*=94.945u + 138.955u = 233.900 u*

^{ }**M**

*2 (*_{0 }n^{1}) =_{Y}

*= 2(1.009u) = 2. 018 u*

*Then:***= [(235.123u + 1.009u) – (233.900 u +2. 018 u)]**

*Q***c**

^{2}

**= [(236.132) - (235.918)] 931MeV/u = 0.214(931MeV/u)**

*Q*
Q = 199.2 MeV

## No comments:

Post a Comment