Question

The half-life period of radium is 1600 years. $2{\text{g}}$ of radium undergoes decay and gets reduced to $0 \cdot 125{\text{g}}$ in:
A) 3200 years
B) 25600 years
C) 8000 years
D) 6400 years

Answer 1

Hint:The half-life period refers to the time taken by a radioactive substance of some amount to reduce to half of its initial amount. So if we can express the ratio of the current amount of radium left to its initial amount as a fraction of $\dfrac{1}{2}$, then we can obtain the number of half-lives taken to reduce to the given amount.

Complete step by step solution.

Step 1: List the given parameters of the radium sample.
The half-life period of radium is given to be ${T_{\dfrac{1}{2}}} = 1600{\text{years}}$.
The initial amount of radium present is given to be $2{\text{g}}$.
The amount of radium present after decay is given to be $0 \cdot 125{\text{g}}$.

Step 2: Express the current amount of radium.
The amount of radium left after a time $t$ is given by, $A = {A_0}{e^{ - \lambda t}}$ ------- (1) where ${A_0}$ is the initial amount of radium and $\lambda $ is the radioactive decay constant.
Now the half-life period of radium is given by, ${T_{\dfrac{1}{2}}} = \dfrac{{\ln 2}}{\lambda }$ -------- (2)
If $n$ is the number of half-lives present in the time $t$ then we have $t = n{T_{\dfrac{1}{2}}}$ -------- (3)
Substituting equation (3) in (A) we get, $A = {A_0}{e^{\left( { - \lambda n{T_{\dfrac{1}{2}}}} \right)}}$ -------- (4)
Substituting equation (2) in (4) we get, $A = {A_0}{e^{ - \left( {\dfrac{{\lambda \times n\ln 2}}{\lambda }} \right)}} = {A_0}{e^{ - n\ln 2}}$ ------- (5)
$ \Rightarrow A = {\left( {\dfrac{1}{2}} \right)^n}{A_0}$
$ \Rightarrow \dfrac{A}{{{A_0}}} = {\left( {\dfrac{1}{2}} \right)^n}$ -------- (6)

Step 3: Substitute values for $A$ and ${A_0}$ in equation (6) to obtain the number of half-lives present in the time $t$.
Substituting for ${A_0} = 2{\text{g}}$ and $A = 0 \cdot 125{\text{g}}$ in equation (6) we get, $\dfrac{{0 \cdot 125}}{2} = {\left( {\dfrac{1}{2}} \right)^n} \Rightarrow n = 4$
Thus the number of half-lives present in the given time is obtained to be $n = 4$.
So substituting for $n = 4$ and ${T_{\dfrac{1}{2}}} = 1600{\text{years}}$ in equation (3) we get, $t = 4 \times 1600 = 6400{\text{years}}$.
$\therefore $ the time taken for radium to reduce to the given amount is obtained to be $t = 6400{\text{years}}$.

Hence the correct option is D.

Note:The radioactive elements suffer an exponential decay. So we express the current amount of radium left by equation (A). The exponential of the natural logarithm will be the argument of the natural logarithm i.e., ${e^{\ln a}} = a$. Also, we have $ - a\ln b = {\left( {\dfrac{1}{b}} \right)^a}$. These two relations are used to simplify equation (5).