Solutions to Problems - Intro. To Nuclear Physics (2)

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  (from Wikipedia table, Googled)

₁₃ Al ²⁷  = M_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) ₈₄ Po 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: _{Z1 +Z2}  X ^{A1 + A2}  =   ₉  X¹⁸  which is 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 e⁰ ®  _Z-1 X ^A   + u

Hence, X = _-1 e⁰

So that: ₄₈ Cd ¹⁰⁹    +_-1 e⁰   ®   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” S _n associated with the neutron.  Estimate the value of this energy if the atomic mass of Ca ⁴³     = 42.98780u, and  Ca ⁴²  = 41.958625u, and take the neutron mass = 1.008665u.

S _n  is just another variant of the Q for a reaction where:

S _n  =   [(total rest mass before decay) –

                                (total rest mass after decay )] c²

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

4. 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 ⁷ (p, n) Be ⁷

Solution:

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

Q = [M_a + M_X  - M_Y – M_b]c²

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 (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:

₉₂U 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

₉₂U ²³⁵   = 235.123u

₄₂ Mo ⁹⁵   =  94.945u

₅₇ La ¹³⁹   =  138.955u

Q = [M_a + M_X  - M_Y – M_b]c²

Where:

₉₂U ²³⁵    = M_a   = 235.123u

₀ n ¹       = M_X  = 1.009u

₄₂ Mo ⁹⁵  +  ₅₇ La ¹³⁹  = M_b   =94.945u + 138.955u =  233.900 u

 2 (₀ n ¹)  = M_Y  =   2(1.009u) =  2. 018 u

Then:

Q = [(235.123u + 1.009u) – (233.900 u +2. 018 u)] c²

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

Q =  199.2 MeV