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Physics Question on Nuclei

In a radioactive material the activity at time t1 is R1 and at a later time t2, it is R2. If the decay constant of the material is λ\lambda, then :

A

R1 = R2 e−λ(t1−t2)\text{e}^{-\lambda(t_1-t_2)}

B

R1= R2 eλ(t1−t2)\text{e}^{\lambda(t_1-t_2)}

C

R1= R2 e(t2t1)\text{e}^{\bigg(\frac{t_2}{t_1}\bigg)}

D

R1 = R2

Answer

R1 = R2 e−λ(t1−t2)\text{e}^{-\lambda(t_1-t_2)}

Explanation

Solution

Radioactive Decay Constant: The radioactive decay of a material follows the formula: N(t) = N0 ×\times e(−λt)\text{e}^{(-\lambda t)}, where N(t) is the activity at time t, N0 is the initial activity,λ\lambda is the decay constant, and t is time.
If R1 is the activity at time t1 and R2 is the activity at time t2, we have:
R1 = N0 ×\times e(−λt1)\text{e}^{(-\lambda t_1)}
R2 = N0 ×\times e(−λt2)\text{e}^{(-\lambda t_2)}
Dividing these equations: R1R2\frac{\text{R}_1}{\text{R}_2} = e(−λt1)e(−λt2)\frac{\text{e}^{(-\lambda t_1)}}{\text{e}^{(-\lambda t_2)}}= e(−λ(t1−t2))\text{e}^{(-\lambda(t_1-t_2))}

So, the correct option is R1 = R2 e−λ(t1−t2)\text{e}^{-\lambda(t_1-t_2)}