Answers from problem # 16

(a) Given;
Elements X and Y
Mass numbers Mx=206 and My=208.
Atomic numbers Zx=Zy≡Z
Neutrons of X,Nx=124.
Required;  Neutrons in Y and its atomic number,Zy. 
Solution.
The reader should understand that, Since X and Y have similar atomic numbers and so is their protons.
Then the two are isotopes.
Thus; proton numbers are the same except for neutrons.
Recall: Mass number=atomic number/proton number+Neutrons number.
i.e. Mx=Zx+Nx.
Thus; Zx=Mx-Nx.
⇒Zx=206 - 124.
Thus; Zx=82.
.: The atomic number of X=82.

Since; Zx=Zy≡Z=82.
.: The atomic number of Y also is=82.

From the eqn:
Ny=My-Zy.
Where all symbols carry their usual meanings.
⇒Ny=208-82.
Thus; Ny=126.
.: The number of neutrons in Y=126.

(b) Given:
Europium isotopes,
151
    Eu of mass ,m₁=150.91976 a.m.u and
153
    Eu of mass,m₂=152.9196 a.m.u.
Average atomic mass of Eu, say A.A.M =151.96 a.m.u
Required;
% relative abundances for each isotope.
Solution:
A.A.M= sum of products of isotopic abundance and their atomic masses divided by 100
Now case I:
      151
For    Eu isotope m₁=150.91976 a.m.u and
      153
For    Eu isotope m₂=152.9196 a.m.u
Let: X₁ and X₂ be their respective abundances.
Clearly
(m₁•X₁ + m₂•X₂)= A.A.M
           100
m₁•X₁ + m₂•X₂ = 100•A.A.M
⇒150.91976X₁+152.9196X₂=15196...(i)
Then case II.
Sum of Relative abundances=100%.
X₁ + X₂ = 100..................(ii)

Solving equations (i) & (ii) simultaneously gives
X₁ = 47.98% and
X₂ = 52.02%.
.: The % abundances for each isotopes are:
47.98% and 52.02% 
       151              153
For    Eu  and    Eu respectively.

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