Chapter 19, Problem 54A

Interpretation Introduction

Interpretation: The correct statement needs to be determined.

Concept introduction: During radioactive decay, instantaneous conversion of all atoms of one element into atoms of another element does not occur but the decay process takes time.

The half-life of a nucleus can be defined as the time taken for one-half of the original sample of nuclei to decay.

Half life can be calculated from the decay constant of the radionuclide. The relationship between them is,

  $\begin{array}{l}{t}_{1/2}=\frac{\text{ln2}}{\lambda }\text{ where, }{t}_{1/2}=\text{ half life of the radionuclide}\\ \lambda \text{= decay constant of the process}\end{array}$

The n/p ratio of an atomic nucleus is the ratio of its number of the neutron to its number of protons.

n/p ratio indicates the stability of the nucleus. A nuclide is most stable when n/p ratio is equal to 1 (when number of neutrons and number of protons are same).

Answer to Problem 54A

a, b and d are correct options.

Explanation of Solution

Shorter half life means it takes less time to decay to half of its amount. This means that the nuclide is unstable compared to a nuclide which has higher value of half life.

In radioactive decay process an unstable atomic nucleus loses energy by emitting ionizing particles and radiation. Some mass of the product is loss which gets converted into energy by the following equation,

  $\begin{array}{l}E=\Delta m{c}^2\Delta m=\text{ difference between mass of parent nuclide and daughter nuclide}\\ c=\text{ speed of light}\end{array}$

If n/p ratio is greater or lesser than 1 then radioactive nuclide decays by various radioactive decay process to approach a value which closer to one or equal to one. So, decay continues until a stable nuclide is formed.

Conclusion

Nuclear fusion is the process by which two or lighter nuclei fuse together to form a heavier nucleus with the release of large amounts of energy.

Example of nuclear fusion:

  ${}_{\text{ 1}}^{\text{ 2}}H+{}_{\text{ 1}}^{\text{ 3}}H\to {}_2^{4}He+{}_0^{\text{1}}n$

Here deuterium and tritium (lighter nucleus) combines to form helium (heavier nucleus).

So, here radionuclide does no decay but it fuses to together to form a different nucleus.

Option c is incorrect.