Question 260: Atoms & Nuclei

Question

Activities of three radioactive substances A, B
and C are represented by the curves A, B and
C, in the figure. Then their half-lives
$${T_{{1 \over 2}}}\left( A \right)$$ : $${T_{{1 \over 2}}}\left( B \right)$$ : $${T_{{1 \over 2}}}\left( C \right)$$ are in the ratio :

Accepted Answer

$$R = {R_0}{e^{ - \lambda t}}$$
$$\ln R = ln{R_0} - \lambda t$$
$${\lambda _A} = {6 \over {10}} \Rightarrow {T_A} = {{10} \over 6}\ln 2$$
$${\lambda _B} = {6 \over 5} \Rightarrow {T_B} = {{5\ln 2} \over 6}$$
$${\lambda _C} = {2 \over 5} \Rightarrow {T_C} = {{5\ln 2} \over 2}$$
$${{10} \over 6}:{5 \over 6}:{{15} \over 6}::2:1:3$$

Answer

3 : 2 : 1

Answer

2 : 1 : 1

Answer

4 : 3 : 1

Answer

2 : 1 : 3

Atoms & Nuclei

Activities of three radioactive substances A, B and C are represented by the curves A, B and C, in the figure. Then their half-lives ${T_{{1 \over 2}}}\left( A \right) : {T_{{1 \over 2}}}\left( B \right) : {T_{{1 \over 2}}}\left( C \right)$ are in the ratio :

Choose the correct answer: