Question

One gram of radium is reduced by $2mg$ in $5$ years by $\alpha -decay$. The half-life of radium is:
$A)1520.4years$
$B)1671.7years$
$C)1617.5years$
$D)1732.86years$

Answer 1

Half-life of a radioactive element is the time required for the element to reduce to half its size, during the process of radioactive decay. The value of the decay constant is found out from the expression of $\alpha -decay$. The half-life of radium is determined using this decay constant.
Formula used:
$1)N={{N}_{0}}{{e}^{-\lambda t}}$
where
$N$ is the final amount of the radioactive element
${{N}_{0}}$ is the initial amount of the radioactive element
$\lambda$ is the decay constant
$t$ is the time required for the amount of radioactive element to become $N$ from ${{N}_{0}}$
$2){{t}_{\dfrac{1}{2}}}=\dfrac{\ln (2)}{\lambda }=\dfrac{0.693}{\lambda }$
where
${{t}_{\dfrac{1}{2}}}$ is the half-life of the radioactive element
$\lambda$ is the decay constant radioactive element

Complete step-by-step solution:
In nuclear physics, the half-life of a radioactive element is the time required for the radioactive element to reduce to half its size, in the process of radioactive decay.
The radioactive decay mentioned in the question is $\alpha -decay$ and the radioactive element considered is radium. We are supposed to find the half-life of radium. We are provided with the results of an experiment, which involved $\alpha -decay$ of radium through a period of $5$ years. The initial amount of radium was $1g$ and the final amount of radium after $5$ years is $2mg$ less than the initial amount. The general expression of $\alpha -decay$ is given by
$N={{N}_{0}}{{e}^{-\lambda t}}$
where
$N$ is the final amount of the radioactive element
${{N}_{0}}$ is the initial amount of the radioactive element
$\lambda$ is the decay constant
$t$ is the time required for the amount of radioactive element to become $N$ from ${{N}_{0}}$

Substituting the values given in our question, we have
$N={{N}_{0}}{{e}^{-\lambda t}}\Rightarrow 998=1000({{e}^{-5\lambda }})$
where
$N$ is the final amount of radium, which is equal to $(1g-2mg)=(1000mg-2mg)=998mg$
${{N}_{0}}$ is the initial amount of radium, which is equal to $1g$ or $1000mg$
$t$ is the time required for the amount of radium to become $998mg$ from $1g$, which is equal to $5$ years
$\lambda$ is the decay constant
Let this be equation 1.
Solving equation 1, we have
$998=1000({{e}^{-5\lambda }})\Rightarrow \dfrac{998}{1000}={{e}^{-5\lambda }}$
Taking natural log$(\ln )$ on both sides, we have
$\ln \left( \dfrac{998}{1000} \right)=\ln ({{e}^{-5\lambda }})=-5\lambda$
Solving this equation further, we have
$\ln \left( 0.998 \right)=-5\lambda \Rightarrow -0.0020=-5\lambda \Rightarrow \lambda =\dfrac{-0.0020}{-5}=4\times {{10}^{-4}}$
Therefore, the decay constant of radium is given by
$\lambda =4\times {{10}^{-4}}$
Let this be equation 2.
Now, let us determine the half-life of radium using the formula given below.
${{t}_{\dfrac{1}{2}}}=\dfrac{\ln (2)}{\lambda }=\dfrac{0.693}{\lambda }$
where
${{t}_{\dfrac{1}{2}}}$ is the half-life of radium
$\lambda$ is the decay constant of radium
Substituting the value of decay constant from equation 2, we have
${{t}_{\dfrac{1}{2}}}=\dfrac{0.693}{\lambda }=\dfrac{0.693}{4\times {{10}^{-4}}}=0.17325\times {{10}^{4}}=1732.5years$
Therefore, the half-life of radium is given by
${{t}_{\dfrac{1}{2}}}=1732.5years$
The option close to this determined value is $D$. Hence, the correct option to be marked is $D$.

Note: Radioactive decay of a radioactive element can also be expressed using the following formula:
$N(t)={{N}_{0}}{{\left( \dfrac{1}{2} \right)}^{\dfrac{t}{{{t}_{\dfrac{1}{2}}}}}}$
where
$N$ is the final amount of the radioactive element
${{N}_{0}}$ is the initial amount of the radioactive element
$t$ is the time required for the amount of radioactive element to become $N$ from ${{N}_{0}}$
${{t}_{\dfrac{1}{2}}}$ is the half-life of the radioactive element
In this case, it can be seen that decay constant of the radioactive element is not involved and students can consider this formula as an easy method to solve the given question.