Question

A radioactive material decays by simultaneous emission of two particles with respective half-lives 1620 and 810 years. The time, in years, after which one-fourth of the material remains is:
A. 1080
B. 2430
C. 3240
D. 4860

Answer 1

Hint: Radioactive processes are generally 1st order reactions. The half-life of the reaction is the time required to reduce half of the concentration of the reactant. The half-life of the 1st order reaction does not depend upon the concentration of the reactant.

Formula used: \[N = {N_0}{\left( {\dfrac{1}{2}} \right)^n}\], \[n = \dfrac{{time{\text{ }}of{\text{ }}deccay}}{{effective{\text{ }}half - life}}\],\[\lambda = \dfrac{{\ln 2}}{T}\] ,\[{\lambda _1} + {\lambda _2} = \dfrac{{\ln 2}}{{{T_1}}} + \dfrac{{\ln 2}}{{{T_2}}}\] ,\[\dfrac{1}{T} = \dfrac{1}{{{T_1}}} + \dfrac{1}{{{T_2}}}\]

Complete step by step answer:
According to the law of Rutherford - Soddy the number of atoms left after n number of Half-life is \[N = {N_0}{\left( {\dfrac{1}{2}} \right)^n}\]. Where \[{{\text{N}}_{\text{o}}}\]is the original number of atoms. The number of half-lives
\[n = \dfrac{{time{\text{ }}of{\text{ }}deccay}}{{effective{\text{ }}half - life}}\] . The relation between effective disintegration constant(\[{\text{\lambda }}\])and half-life(\[{\text{T}}\]) is
\[\lambda = \dfrac{{\ln 2}}{T}\] .
Now, \[{\lambda _1} + {\lambda _2} = \dfrac{{\ln 2}}{{{T_1}}} + \dfrac{{\ln 2}}{{{T_2}}}\]
and the effective half-life is
\[\
\dfrac{1}{T} = \dfrac{1}{{{T_1}}} + \dfrac{1}{{{T_2}}} \\
\dfrac{1}{T} = \dfrac{1}{{1620}} + \dfrac{1}{{810}} \\
\dfrac{1}{T} = \dfrac{{1 + 2}}{{1620}} \\
T = 540 \\
\]
Therefore, the number of half-lives is,
\[\
  n = \dfrac{{time{\text{ }}of{\text{ }}deccay}}{{effective{\text{ }}half - life}} \\
  n = \dfrac{t}{{540}} \\
 \]
Now according to the question, \[N = {N_0}{\left( {\dfrac{1}{2}} \right)^n}\]
Therefore,
\[\
\dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^2} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{540}}}} \\
  {\left( {\dfrac{1}{2}} \right)^2} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{{540}}}} \\
  2 = \dfrac{t}{{540}} \\
  t = 1080 \\
 \]
Therefore, the time in years, after which one-fourth of the material remains is 1080 years.

Note:
Chemical reactions are assigned reaction orders that describe their kinetics. A first-order reaction is a reaction that proceeds at a rate that depends linearly on only one reactant concentration. Further, the half-life of a species is the time it takes for the concentration of that substance to fall to half of its initial value.