Solveeit Logo

Question

Question: The half-life of a radioactive element is 10 years. What percentage of it will decay in 100 years? ...

The half-life of a radioactive element is 10 years. What percentage of it will decay in 100 years?
A) 99.90/099.9{\scriptstyle{}^{0}/{}_{0}}
B)100/010{\scriptstyle{}^{0}/{}_{0}}
C)500/050{\scriptstyle{}^{0}/{}_{0}}
D) 66.50/066.5{\scriptstyle{}^{0}/{}_{0}}

Explanation

Solution

The time of the decay of the substance at the time ‘t’ is related to the number of half-life residing in ‘t’, t=nt1/2\text{t=n}{{\text{t}}_{\text{1/2}}}.The amount of substance not decaying is find out by the equation, [A]=[A0]2n\left[ \text{A} \right]\text{=}\dfrac{\left[ {{\text{A}}_{\text{0}}} \right]}{{{\text{2}}^{\text{n}}}}. The amount of substance decays at time ‘t’ is obtained by considering the total amount of substance as a 1000/0100{\scriptstyle{}^{0}/{}_{0}}

Complete answer:
The half-life t1/2{{\text{t}}_{\text{1/2}}} for a radioactive element is defined as the amount of the time required for a given amount of substance to reduce by the half of its initial amount because of the decay or emission of
We are given with the half-life of radioactive elements t1/2=10 years{{\text{t}}_{\text{1/2}}}=10\text{ years}
The time at which the total amount of substance undergoes the decay,100 years100\text{ years}
Let the initial amount of the radioactive substance be equal to the [A0]\left[ {{\text{A}}_{\text{0}}} \right] at the time t=0\text{t=0}
Consider the number of substance decays at the t=100 years\text{t=100 years} is equal to the[A]\left[ \text{A} \right].
We know that the time required for the decay of radioactive material is equal to the ‘n’ number of times to the half-life of an element.
The formula is as:
t=nt1/2\text{t=n}{{\text{t}}_{\text{1/2}}}
Let’s first find out the total number of half-lives of 10 years present in the time of 100 years.
100=n×10\text{100=n}\times \text{10}
Rearrange the formula for n we get,
n=10010\text{n=}\dfrac{100}{10}
n=10\text{n=}10
Thus, the total number of half-life residing in the time ‘t’ is 10.
We have the other formula which relates the initial amount of radioactive element [A0]\left[ {{\text{A}}_{\text{0}}} \right] is related to the amount of substance at a time ‘t’.
[A]=[A0]2n\left[ \text{A} \right]\text{=}\dfrac{\left[ {{\text{A}}_{\text{0}}} \right]}{{{\text{2}}^{\text{n}}}}
Let us substitute values for ‘n’.
=[A0]210\text{=}\dfrac{\left[ {{\text{A}}_{\text{0}}} \right]}{{{\text{2}}^{\text{10}}}}
=[A0]1024\text{=}\dfrac{\left[ {{\text{A}}_{\text{0}}} \right]}{1024}
Since,210=1024{{2}^{10}}=1024
[A]=[A0]×9.765×104\left[ \text{A} \right]\text{=}\left[ {{\text{A}}_{\text{0}}} \right]\times 9.765\times {{10}^{-4}}
Let's find out the percentages of the radioactive substance undergoes the decay at the 100 years
0/0 radioactive substance decay=100-[A][A0]{\scriptstyle{}^{\text{0}}/{}_{\text{0}}}\text{ radioactive substance decay=100-}\dfrac{\left[ \text{A} \right]}{\left[ {{\text{A}}_{\text{0}}} \right]}
Put the values in the above equation.
=100-[A0]×9.765×104[A0]\text{=100-}\dfrac{\left[ {{\text{A}}_{\text{0}}} \right]\times 9.765\times {{10}^{-4}}}{\left[ {{\text{A}}_{\text{0}}} \right]}
=100-[A0]×9.765×104[A0]\text{=100-}\dfrac{\left[ {{\text{A}}_{\text{0}}} \right]\times 9.765\times {{10}^{-4}}}{\left[ {{\text{A}}_{\text{0}}} \right]}
=100-9.765×1040/0\text{=100-}9.765\times {{10}^{-4}}{\scriptstyle{}^{0}/{}_{0}}
=99.90/0\text{=}99.9{\scriptstyle{}^{0}/{}_{0}}
The amount of radioactive substances decays at the 100 years is equal to the 99.90/099.9{\scriptstyle{}^{0}/{}_{0}}.

Hence, (A) is the correct option.

Note:
Here we are asked to find out the amount of substance decays at the time t hence always subtract the number of substance decays from the 100 to get the desired answer.