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

Hint: 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.