Solutions To Nuclear Fission Problems (Nuclear Physics, Part 3)

 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]c²

Where:

₀ n ¹ = M_b     = 1.008665u

a = M_X  =  ₂ He ⁴  =  4.002603u

₁₅ P ³⁰    = M_Y  =   29.9783138u 

₁₃ Al ²⁷  = M_a   =  26.9815386u

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) ₈₄ Po ²¹⁵   ®  X + a

 

b)   N ¹⁴ (a, X)  O ¹⁷

 

c) ₄₈ Cd ¹⁰⁹    +   X ®   ₄₇  Ag ¹⁰⁹     +  u

Solutions:

T   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: 

_{Z1 +Z2}  X ^{A1 + A2}  =   ₉  X ¹⁸  

Which is fluorine,  or X = F (e.g. ₉ F ¹⁸ )

c)   c) 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 e⁰ ®  _Z-1 X ^A   + u

Hence, X = _-1 e ⁰

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

 

a) H ² (d, p)  H ³    (Note: d is for deuterium or ₁ H ² )

 

b) Li ⁶ (p, a) H ³

 

c) Li ⁷ (p, n) Be ⁷

d) F ¹⁹ (p, a) O ¹⁶

Solutions:

a)  ₁H ²  +   ₁H ²  ®  ₁H ¹  +   ₁H ³  

Q = [M_a + M_X  - M_Y – M_b]c²

 Then:

   ₁ H ² = M_a     = 2.01410 u

₁ H ²  = M_X  =  2.01410 u

₁H ¹ = M_b   = 1.007825 u

   ₁ H ²  = M_Y  =   3.01604 u

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 )

Q  = 4.09 MeV

Solutions to (b) - (d) follow analogous approaches,  simply substituting for the respective  M_a ,  M_X , M_Y, M_b  in the specific nuclear reactions.  E.g. for (c):

Li ⁷ (p, n) Be ⁷

For:   ₃ Li ⁷   +   ₁H ¹  ®    ₀ n ¹  +   ₄ Be ⁷  

Q = [M_a + M_X  - M_Y – M_b]c²

M_a  =  7.016004 u  (for ₃  Li  ⁷  )

M_X  = 1.007825 u  (for  ₁H ¹ )

M_Y =  1.009u    (for   ₀ n ¹  )

M_b  =  7.0169287 u  (for   ₄ Be ⁷ )

Q = [7.016004 u + 1.007825 u  - 1.009u – 7.0169287 u] c²

Q= - 0.004u (931 Mev/u )  =  -3.72 MeV