Answer
Verified
430.2k+ views
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).
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).
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Which are the Top 10 Largest Countries of the World?
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Difference Between Plant Cell and Animal Cell
Give 10 examples for herbs , shrubs , climbers , creepers
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you graph the function fx 4x class 9 maths CBSE
Write a letter to the principal requesting him to grant class 10 english CBSE