Question

Calculate the time interval between $33\%$ decay and $67\%$ decay if half-life of a substance is $20$ minutes.

• A. $40$ minutes
• B. $20$ minutes
• C. $60$ minutes
• D. $13$ minutes

Answer 1

The correct option is B

$20$ minutes

The above question can be solved as:

Step 1:

Half life a substance $=20$minutes

Step 2:

Rate of decay at time $\mathrm{t}$ :

${\mathrm{N}}_{\mathrm{t}}={\mathrm{N}}_0{\mathrm{e}}^{-\mathrm{\lambda t}}$

$\Rightarrow \mathrm{\lambda t}=\ln ({\mathrm{N}}_{\mathrm{o}}/{\mathrm{N}}_{\mathrm{t}})--------\left(\mathrm{i}\right)$

Step 3:

Applying the formula of equation ($\mathrm{i}$i) for $33\%$ decay and (iii) for $67\%$ decay:

$$\begin{gathered} {\mathrm{\lambda t}}_2=\ln (100/100-67) \\ {\mathrm{\lambda t}}_2=\ln (100/33)--------\left(\mathrm{ii}\right) \end{gathered}$$

$$\begin{gathered} {\mathrm{\lambda t}}_1=\ln (100/100-33) \\ {\mathrm{\lambda t}}_1=\ln (100/67)--------\left(\mathrm{iii}\right) \end{gathered}$$

Equation($\mathrm{ii}$) $-$ Equation($\mathrm{iii}$)

$$\begin{gathered} \Rightarrow \mathrm{\lambda }({\mathrm{t}}_2–{\mathrm{t}}_1)=\ln (0.67/0.33) \\ =\ln (2.03)\cong \ln \left(2\right) \\ \Rightarrow {\mathrm{t}}_2–{\mathrm{t}}_1=\ln 2/\mathrm{\lambda } \\ \Rightarrow \ln 2/\mathrm{\lambda }=0.693/\mathrm{\lambda } \\ \Rightarrow 0.693/\mathrm{\lambda }={\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ \Rightarrow {\mathrm{t}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=20\min . \\ \Rightarrow {\mathrm{t}}_2–{\mathrm{t}}_1=20\min . \end{gathered}$$

Therefore, the time interval between $33\%$ decay and $67\%$ decay is $20$ minutes.