Answer to Question #160166 in Atomic and Nuclear Physics for Terry

Question

Two samples of the same radioactive nuclide are prepared. Sample G hadms twice the initial activity of sample H. How does the half-life of G compare with the half-life of H?
(A) it is two times larger than that of G
(B) it is thev same as that of G
(C) it is half of that of G
(D) it is larger than that of G 4 times.

Answer 1

The relationship between the activity $R$ and the number of nuclei in the sample, $N$, can be written as follows:

$R=N\lambda,$

here, $\lambda$ is decay constant.

Since, both samples have the same size (and, therefore, have the same initial number of nuclei), we can write:

$R_G=N_o\lambda_G,$ $R_H=N_o\lambda_H.$

Dividing $R_G$ by $R_H$, we get:

$\dfrac{R_G}{R_H}=\dfrac{\lambda_G}{\lambda_H},$ $\dfrac{2R_H}{R_H}=\dfrac{\lambda_G}{\lambda_H},$ $\lambda_G=2\lambda_H.$

The relationship between the half-life of a sample, $T_{1/2}$, and its decay constant $\lambda$ can be written as follows:

$T_{1/2}=\dfrac{ln2}{\lambda}.$

Then, we can write the half-life of each samples as follows:

$T_{1/2,G}=\dfrac{ln2}{\lambda_G},$ $T_{1/2,H}=\dfrac{ln2}{\lambda_H}.$

Dividing $T_{1/2,G}$ by $T_{1/2,H}$ and substituting $\lambda_G$ we get:

$\dfrac{T_{1/2,G}}{T_{1/2,H}}=\dfrac{\lambda_H}{\lambda_G}=\dfrac{\lambda_H}{2\lambda_H}=\dfrac{1}{2},$ $T_{1/2,H}=2T_{1/2,G}.$

As we can see from calculations, the half-life of sample H is two times larger than that of G. Therefore, the correct answer is (A).

Answer:

(A) it is two times larger than that of G.