In a radioactive material the activity at time t_1 is R_1 and at a later time t_2 , it is R_2 . If the decay constant of the material is λ , then:

Correct answer: B. (R_1=R_2 e^ -λ(t_1-t_2) ) R_1=R_0 e^ -λ_4 and R_2=R_0 e^ -λ λ_2 aligned (R_1)/(R,)& = e^ -λ λ_1 e^ -λ λ_2 & =e^ -λ ( t_1-t_2) R_1 & =R_2 e^ -λ( t_1-t_2) aligned

NEET 2006

Exam: NEET
Year: 2006
Subject: Physics
Chapter: Nuclear Physics

NEET 2006 · Physics · Nuclear Physics

In a radioactive material the activity at time t_1 is R_1 and at a...

In a radioactive material the activity at time

t_1

R_1

and at a later time

t_2

, it is

R_2

. If the decay constant of the material is

\lambda

, then:

Options

A $R_1=R_2$
B(R_1=R_2 e^{-\lambda\left(t_1-t_2\right)})
C(R_1=R_2 e^{\lambda\left(t_1-t_2\right)})
D(\mathrm{R}_1=\mathrm{R}_2\left(\mathrm{t}_2 / \mathrm{t}_1\right))

Correct answer

B. (R_1=R_2 e^{-\lambda\left(t_1-t_2\right)})

Step-by-step solution

R_1=R_0 e^{-\lambda_4} and

R_2=R_0 e^{-\lambda \lambda_2}

\begin{aligned} \Rightarrow \quad\frac{R_1}{R,}& =\frac{e^{-\lambda \lambda_1}}{e^{-\lambda \lambda_2}}\\ & =e^{-\lambda \left( t_1-t_2\right)} \\ \Rightarrow \quad R_1 & =R_2 e^{-\lambda\left( t_1-t_2\right)} \end{aligned}

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