Question 11.1: Write a balanced nuclear equation for the decay of each of t......

In each case, the atomic and mass numbers of the daughter nucleus are obtained by writing the symbols of the parent nucleus and the particle emitted by the nucleus (alpha or beta). Then the equation is balanced.
a. Let X represent the daughter nuclide, the product of the radioactive decay. Then
^{70}_{31}Ga  →  ^{      0}_{-1}\beta + X

The sum of the superscripts on both sides of the equation must be equal, so the superscript for X must be 70. In order for the sum of the subscripts on both sides of the equation to be equal, the subscript for X must be 32. Then 31 = (-1) + (32). As soon as we determine the subscript of X, we can obtain the identity of X by looking at a periodic table. The element with an atomic number of 32 is Ge (germanium). Therefore,

^{70}_{31}Ga  →  ^{      0}_{-1}\beta  +  ^{70}_{32}Ge
b. Letting X represent the daughter nuclide, we have, for the alpha decay of ^{144}_{    60}Nd,
^{144}_{    60}Nd → ^{4}_{2}α + X
We balance the equation by making the superscripts on each side of the equation total 144 and the subscripts total 60. We get
^{144}_{    60}Nd → ^{4}_{2}α + ^{140}_{    58}Ce
c. Similarly, we write
^{248}_{100}Fm → ^{4}_{2}α + X
Balancing superscripts and subscripts, we get
^{248}_{100}Fm → ^{4}_{2}α + ^{244}_{    98}Cf
In alpha emission, the atomic number of the daughter nuclide always decreases by 2, and the mass number of the daughter nuclide always decreases by 4.
d. Finally, we write
^{113}_{    47}Ag → ^{      0}_{-1}\beta + X
In beta emission, the atomic number of the daughter nuclide always increases by 1, and the mass number does not change from that of the parent. The balancing procedure gives us the result
^{113}_{    47}Ag → ^{      0}_{-1}\beta + ^{113}_{    48}Cd