Question

At a given instant, say t=0, two radioactive substances A and B have equal activities. The ratio $\dfrac{{{{\text{R}}_{\text{B}}}}}{{{{\text{R}}_{\text{A}}}}}$ of their activities after time t itself decays with time as ${e^{ - 3t}}$ . If the half life of A is $m_2$ , the half life of B is:
A) $\dfrac{{\ln 2}}{2}$
B) $2\ln 2$
C) $\dfrac{{\ln 2}}{4}$
D) $4\ln 2$

Answer 1

Hint:Here we will be using the concept of half life of a radioactive substance that is time required to reduce half of the original substance.

Complete step-by-step answer:
Let us consider the half life of the radioactive substance is defined as the time required for one half of the radioactive substance to disintegrate. It is denoted by ${T_{1/2}}$
We know that a radioactive substance has ${N_0}$ atoms at time t=0. Then the number of atoms remained undecayed.
After time t is N . that is we have the $N = {N_0}{e^{ - \lambda t}}$
When we have the t= ${T_{1/2}}$, the value of N is given as $N = \dfrac{{{N_0}}}{2}$
Therefore, we can say that
$\dfrac{{{N_0}}}{2} = {N_0}{e^{ - \lambda {T_{1/2}}}}$
Now let us take logs on both sides.
Taking log on both sides, we have $\lambda {T_{1/2}} = {\log _e}2 = 2.303 \times {\log _{10}}2 = 0.693$
Therefore, the time period is given as ${T_{1/2}} = \dfrac{{0.693}}{\lambda }$
Also we can write the time period as, ${T_{1/2}} = \dfrac{{\ln 2}}{\lambda }$
Given, let us consider the half life of A=$m_2$
${T_{1/2}} = {m_2}$
$\dfrac{{\ln 2}}{{{\lambda _A}}} = {m_2}$
${\lambda _A} = \dfrac{{\ln 2}}{{{m_2}}}$
We can consider that At t=0, ${R_A} = {R_B}$
${N_A}{e^{ - {\lambda _A}t}} = {N_B}{e^{ - {\lambda _B}t}}$
${N_A} = {N_B}$ Since t=0
After some time t we have,
$\dfrac{{{R_B}}}{{{R_A}}} = \dfrac{{{e^{ - {\lambda _B}t}}}}{{{e^{ - {\lambda _A}t}}}} = {e^{ - 3t}}$
${e^{ - \left( {{\lambda _B} - {\lambda _A}} \right)t}} = {e^{ - 3t}}$
${\lambda _B} = 3 + {\lambda _A}$
${\lambda _B} = 3 + \dfrac{{\ln 2}}{{{m_2}}}$
Therefore, the half period is given as ${T_{1/2}} = \dfrac{{\ln 2}}{{\left( {3 + \dfrac{{\ln 2}}{{{m_2}}}} \right)}}$
Incase ${m_2} = \ln 2$ then the half period is given as
${T_{1/2}} = \dfrac{{\ln 2}}{4}$
Therefore the half life period of the B is ${T_{1/2}} = \dfrac{{\ln 2}}{4}$
Hence the correct option is (C)

Additional information:
Radioactivity was discovered by Henry Becquerel in 1896. He found that Uranium salt wrapped up in paper emitted certain penetrating radiation which affected a photographic plate. From this, it observed that the nuclei of all heavy elements with Z above 82 are unstable. The unstable nuclei reduce their energy and become more stable by emitting certain radiation. This phenomenon of emitting certain radiation by heavy elements is called radioactivity.
The process of spontaneous disintegration of the nuclei of elements with the emission of a certain type of radiation is called Radioactivity.
Rutherford and Soddy discovered experimentally that an atom of a radioactive element disintegrates continuously giving rise to the new element. The number of an atom of the original element decreases continuously with time. Then the rate of disintegration also decreases.
According, to radioactive decay law, the rate of disintegration of atoms at any instant of time is directly proportional to the number of radioactive atoms present at that instant.
Let, ${N_0}$ be the number of atoms present initially in the radioactive sample. After time t, the number of radioactive atoms left undecayed in the sample is N. according to decay law,
$\dfrac{{dN}}{{dt}}\propto N$
$\therefore \dfrac{{dN}}{{dt}} = - \lambda N$
Here $\lambda $ is a constant of proportionality and is called decay constant or disintegration constant. The negative sign shows that N is decreasing with time.

Note:The time taken for the concentration of the reactant to reach the 50% of its initial concentration is known as the half-life period.