Question

Two radioactive substances $A$ and $B$ have their half life period $1$ year and $2$ year respectively. Samples $A$ and $B$ initially contain equal number of atoms. After $4$ years the ratio of the number of undecayed atoms of $A$ to the number of undecayed atoms of $B$ of

• A. $1/4$
• B. $1/2$
• C. $2$
• D. $4$
• E. $1$

Answer 1

Answer: (A)

$1/4$

$\begin{array}{c}\text{Given,}\\ \text{two Radioactive atom}\\ \text{- A has half life = 1 year}\\ \text{- B has half life =}2\text{year}\end{array}$

$\begin{array}{c}\text{initially both have equal no of atom}\\ \text{let say}=N_o\end{array}$

$\begin{array}{c}\text{Now,}\\ N_t=N_0{e}^{-\lambda t}\text{where}\\ N_t=N_0\text{of atom left}\\ B_0=\text{initial no of atom}\\ \lambda =\text{decay constant}\end{array}$

$\begin{array}{c}\text{taking log on both side}\\ \log \left(\frac{N_t}{N_0}\right)=\log \left(e^{-\lambda t}\right)\\ \log \left(\frac{N_0}{Nt}\right)=\lambda t\\ \therefore \lambda =\frac{\log 2}{t_{1/2}}\\ N_t=N_0{\left(\frac{1}{2}\right)}^{\frac{t}{T_{1/2}}}\end{array}$

$\begin{array}{c}\text{For}A\\ t=4\text{year}\\ T_{1/2}=\text{1 year}\\ N_t=N_0{\left(\frac{1}{2}\right)}^4\end{array}$

$\begin{array}{c}\text{For}B\\ t=4\text{year}\\ T_{1/2}=2\text{year}\\ N_t=N_0{\left(\frac{1}{2}\right)}^2\end{array}$

$\begin{array}{c}\begin{array}{cc}\text{Now,}& \\ \frac{{\left(N_t\right)}_A}{(Nt{)}_B}& =\frac{{\left(\frac{1}{2}\right)}^4}{{\left(\frac{1}{2}\right)}^2}\\ & =\frac{1}{4}\end{array}\end{array}$