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Half lives of two radioactive elements $A$ and $B$ are ${\text{20}}\,\,{\text{minutes}}$ and ${\text{40}}\,\,{\text{minutes}}$ respectively. Initially the samples have an equal number of nuclei after ${\text{80}}\,\,{\text{minutes}}$ the ratio of decayed numbers $A$ and $B$ nuclei will be.

A. $1:16$

B. $4:1$

C. $1:4$

D. $5:4$

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Last updated date: 25th Jun 2024
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Answer
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Hint: For this type of question we first have to use the formula of half-life which is given as, \[{N_A} = {N_o}{\left( {1/2} \right)^{t/T}}\] . After that find the half-lives of each element individually by putting the values which are given in the question. By this we will get the value of \[{N_A}\] and \[{N_B}\] . According to the question we have to find their ratio. So we just have to divide \[{N_A}\] and \[{N_B}\] . By this method we will find the required ratio.


Complete answer:

Given that there are two radioactive elements $A$ and $B$ which having half live of ${\text{20}}\,\,{\text{minutes}}$ and ${\text{40}}\,\,{\text{minutes}}$ respectively. Also, initially they are having equal numbers of nuclei.We have to find the ratio of decayed numbers of $A$ and $B$ nuclei.Now, with the help of using the formula of half nuclei,

\[N = {N_o}{\left( {1/2} \right)^{t/T}}\] 

We can calculate the half-life value for element $A$


Now, for element $A$

\[{N_A} = {N_o}{\left( {1/2} \right)^{t/T}}\] and we will name it equation $1$

Now, putting the value of variables in the equation $1$ as $t = 80$ and $T = 20$

\[ \Rightarrow {N_A} = {N_o}{\left( {1/2} \right)^{80/20}}\]

Now on solving the above equation we get

\[ \Rightarrow {N_A} = {N_o}{\left( {1/2} \right)^{4}}\]

And further solving more, we get

\[ \Rightarrow {N_A} = {N_o}/16\] 

Decayed number of element $A$ =${N_o} - {N_o}/16 = \dfrac{15}{16} \,{N_o}$....and we will name it equation $2$.


Now, similarly for element $B$ putting the values in the equation $1$ as $t = 80$ and $T = 40$ , we get

\[ \Rightarrow {N_B} = {N_o}{\left( {1/2} \right)^{80/40}}\]

And on solving it, we get

\[ \Rightarrow {N_B} = {N_o}{\left( {1/2} \right)^2}\]

And further solving more, we get

\[ \Rightarrow {N_B} = {N_o}/4\] 

Decayed number of element $B$ =${N_o} - {N_o}/4 = \dfrac{3}{4} \,{N_o}$......and we will name it equation $3$

Now, dividing equation $2$ and $3$ we will get

\[ \Rightarrow {N_A}/{N_B} = \left( {15{N_o}/16} \right)/\left( {3{N_o}/4} \right)\]

And

\[ \Rightarrow {N_A}/{N_B} = \dfrac{5}{4}\]

The ratio of \[A:B\] will be \[5:4\]


Therefore, the correct option is $\left( D \right)$ .


Note: Remember the above equation \[{N_A} = {N_o}{(1/2)^{t/T}}\] to solve these kinds of questions which will be very helpful in future problem solving. Also, these kinds of questions can be very confusing. So, try not to get confused when putting the values in the equation because that always follows step by step which will help to solve questions without getting much time with correct answers.