Question

Half-life of a radioactive substance A is two times the half life of another radioactive substance B. Initially, the number of nuclei of A and B are ${N_A}$ and ${N_B}$ respectively. After three half lives of A, the number of nuclei of both become equal. The ratio of $\dfrac{{{N_A}}}{{{N_B}}}$ will be
A.$\dfrac{1}{2}$
B.$\dfrac{1}{8}$
C.$\dfrac{1}{3}$
D.$\dfrac{1}{6}$

Answer 1

The nuclei of some substances can undergo decomposition or radioactive decay to form a stable nucleus. The number of nuclei remaining after passing through the number of half-lives will have the formula. Given that the half life of A is double to the half-life of B the ratio of number of nuclei will be calculated from the number of half-lives passed.

Complete answer:
Radioactive nucleus is the nucleus that can undergo decay or decomposition to form a stable nucleus. The time required to decay the half of the amount can be called half-life. The number of nucleus that can be remained after passing through number of half-lives will be
$N = {N_0}{\left( {\dfrac{1}{2}} \right)^n}$
Where n is the number of half-lives passed.
The remaining amount of nucleus $A = {N_A}{\left( {\dfrac{1}{2}} \right)^3}$
The remaining amount of nucleus $B = {N_B}{\left( {\dfrac{1}{2}} \right)^6}$
Given that the half-life of nucleus is three and double to the half-life of nucleus B. Thus, nucleus B passed through six half lives and both are equal.
${N_A}{\left( {\dfrac{1}{2}} \right)^3} = {N_B}{\left( {\dfrac{1}{2}} \right)^6}$
By simplification,
$\dfrac{{{N_A}}}{{{N_B}}} = \dfrac{8}{{64}} = \dfrac{1}{8}$
The ratio of $\dfrac{{{N_A}}}{{{N_B}}}$ will be $\dfrac{1}{8}$

Therefore, Option (B) is the correct one.

Note:
The number of half-lives must be known correctly and the nucleus A passes through three half-lives only and the nucleus B passes through double the half-lives of A. Thus, the half-lives must be written and equated as given. The number of nucleus remaining after decay should be expressed as the number of half-lives.