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Writing Nuclear Equations

>> Parts of this equation/concept include:

A -	Predicting Nuclear Decays

Balancing nuclear equations requires that the sum of the mass numbers be the same on each side of the equation. The same is required for the sum of the atomic numbers. The elemental symbol depends on the atomic number. It is normally easier to figure out the missing pieces in atomic number and mass number and then match the symbol to the atomic number.

It is not necessary for all the information to be included in the problem. For example, the atomic and mass number of an particle is consistent, so it may not be included. Since the elemental symbol depends on the atomic number, the atomic number is not always included either. It will probably be easier to do the problem if you fill that information in.

>> Example 1

Fill in the blanks of the following nuclear equations.

1. ²³⁸U + ___
2. n + ¹H ___
3. ¹⁷O + ¹⁴C + ___

Solution:

1. 23892U      42 + 23490Th

   Filling in the atomic number of uranium and both the atomic and mass numbers for the particle, the atomic and mass numbers for the blank can be calculated.

   238 = 4 + A, then A = 234

   92 = 2 + Z, then Z = 90

   Since 90 is the atomic number for thorium, the symbol Th is used.

2. 10n + 11H 21H

   Filling in the appropriate values, the equation for mass number is

   1 + 1 = A, therefore A = 2

   The equation for atomic number is

   0 + 1 = Z; therefore Z = 1, which is the atomic number of hydrogen.

3. 178O +   0–1   146O +  31H

   For mass number, 17 + 0 = 14 + A, so A = 3.

   For atomic number, 8 – 1 = 6 + Z, so Z = 1, and that is the atomic number of hydrogen.

There are four types of nuclear decay. Alpha decay predominates in isotopes with Z > 83. For isotopes with Z < 83, decay will occur if it is outside the belt of stability. You can refer to Figure 2.9 to see if an isotope is within the belt of stability. You may also consider the mass given on the periodic table. Since the mass there is the average mass of all stable isotopes, masses significantly lower than the average mass are neutron-poor and masses significantly above that average mass are neutron-rich. Isotopes that are neutron-rich will undergo -decay. Isotopes that are neutron-poor will decay by electron capture or positron decay. Lighter isotopes (Z < 50) are more likely to decay by positron emission. Heavier elements are more likely to decay by electron capture.

The decay processes start with the isotope and produce the particle (appropriate to the type of decay) and whatever isotope is appropriate to complete the equation. The exception is electron capture, where an electron (the relevant particle) is a reactant rather than a product.

>> Example 2

Write the nuclear decay reactions for the following isotopes.

1. ²³Al
2. ²⁴¹Am
3. ¹⁴⁰Cs
4. ¹⁴⁵Eu
5. ¹⁷⁹W

Solution:

1. The aluminum atom has an average atomic mass of about 27. Therefore the ²³Al isotope is neutron-poor. It is also fairly light. Consequently, the predicted mode of decay is a positron emission.

   2313Al    01 + 2312Mg
   
   

2. The americium isotope has Z = 95; consequently, it will undergo decay.

   24195Am    42 + 23793Np
   
   

3. The cesium atom has an average atomic mass of about 133. Consequently, the cesium-140 isotope is neutron-rich. Therefore it will undergo decay.

   14055Cs     0–1 +   14056Ba
   
   

4. The europium atom has an average atomic mass of about 152. Therefore the europium-145 isotope is neutron-poor. As a heavier element, it is more likely to decay by electron capture than by positron emission.

   19563Eu +    0–1e    19562Sm
   
   

5. Tungsten atoms have an average atomic mass of about 184 . Consequently, the tungsten-179 isotope is neutron-poor. As a heavier atom, it is likely to decay by electron capture.

   17974W +    0–1e    17973Ta
   
   

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