Question

The radioactivity of a sample is $R_1$ at a time $T_1$ and $R_2$ at a time $T_2$. If the half life of the specimen is T, the number of atoms that have disintegrated in the time ($T_2-T_1$) is proportional to

• A. ($R_2T_2-R_1T_1$)
• B. ($R_2-R_1$)
• C. ($R_2-R_1$)/T
• D. ($R_2-R_1$) T/0.693

Answer 1

Answer: (D)

($R_2-R_1$) T/0.693

Answer 2

For a radioactive sample,

$N = \frac{R}{\lambda}$ and $\lambda = \frac{\ln2}{T}$

The number of atoms disintegrated between $T_1$ and $T_2$ is

$\Delta N = N_1 - N_2 = \frac{R_1-R_2}{\lambda} = \frac{(R_1-R_2)T}{\ln2}$

Since $R_1 > R_2$, writing it as $\frac{(R_2-R_1)T}{- \ln2}$ shows proportionality to $(R_2-R_1)T/0.693$.