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Question: In a radioactive material the activity at time \( {t_1} \) is \( {R_1} \) and at a later time \( {t_...

In a radioactive material the activity at time t1{t_1} is R1{R_1} and at a later time t2{t_2} , it is R2{R_2} . If the decay constant of the material is λ\lambda , then:
(A) R1=R2e−λ(t1−t2){R_1} = {R_2}{e^{ - \lambda \left( {{t_1} - {t_2}} \right)}}
(B) R1=R2eλ(t1−t2){R_1} = {R_2}{e^{\lambda \left( {{t_1} - {t_2}} \right)}}
(C) R1=R2=(t2−t1){R_1} = {R_2} = \left( {{t_2} - {t_1}} \right)
(D) R1=R2{R_1} = {R_2}

Explanation

Solution

Hint
The radioactivity of a radioactivity is proportional to the number of nuclei in the sample. If we are measuring the activity we are measuring the number of nuclei that are disintegrating. With time the number of nuclei present in the sample decreases.

Complete step by step answer
The unit of activity is Becquerel (Bq). One Becquerel is the decay rate of one disintegration per one second.
The decay rate or rate of disintegration is the number of decays per second. The expression for the rate of disintegration is given as,
R=−dNdtR = - \dfrac{{dN}}{{dt}}
Where, NN is the number of nuclei in the sample.
And we have the rate of number of nuclei per second is proportional to the number of nuclei. Therefore,
−dNdt=λN- \dfrac{{dN}}{{dt}} = \lambda N
Where, λ\lambda is the decay constant.
Comparing the both equations, we get
R=λNR = \lambda N
We have N=N0e−λtN = {N_0}{e^{ - \lambda t}}
Where, NN is the number of nuclei at time tt and N0{N_0} is the number of nuclei at t=0t = 0 .
Substituting this in the equation for activity gives,
R=λN0e−λtR = \lambda {N_0}{e^{ - \lambda t}}
=R0e−λt= {R_0}{e^{ - \lambda t}}
Where, λN0=R0\lambda {N_0} = {R_0} . And R0{R_0} is the activity at time t=0t = 0 .
Therefore, R=R0e−λtR = {R_0}{e^{ - \lambda t}}
For time t1{t_1} , R1=R0e−λt1...........(1){R_1} = {R_0}{e^{ - \lambda {t_1}}}...........\left( 1 \right)
And for time t2{t_2} , R2=R0e−λt2...............(2){R_2} = {R_0}{e^{ - \lambda {t_2}}}...............\left( 2 \right)
Dividing equation (1)\left( 1 \right) and (2)\left( 2 \right) , we get
R1R2=R0e−λt1R0e−λt2\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{{R_0}{e^{ - \lambda {t_1}}}}}{{{R_0}{e^{ - \lambda {t_2}}}}}
R1R2=e−λ(t1−t2)\dfrac{{{R_1}}}{{{R_2}}} = {e^{ - \lambda \left( {{t_1} - {t_2}} \right)}}
R1=R2e−λ(t1−t2){R_1} = {R_2}{e^{ - \lambda \left( {{t_1} - {t_2}} \right)}}
The answer is option A.

Note
If the radioactivity is greater means shorter will be the half -life of the material. The activity can be varied according to the time. That is the radiation emitted will vary in time.