The half-life of Pb-210 is 22 years. If 2 grams of Pb-210 is let to decay, what will be its amount after 11 years?
A: 0.1414 grams.
B: 1.414 grams.
C: 2.828 grams
D: 0.707 grams.
Answer
652.5k+ views
Hint: The half-life of the substance is the time taken by it to decay up to half of its initial concentration. The half-life will tell you about the rate constant for the reaction.
Complete step by step answer:
We know that, the reaction of Pb-210 is as follows;
$^{210}Pb \to X$
Hence the kinetics can be shown as given in the table below;
Now as we know the half-life we can find out the rate constant
Since the decay follows first order kinetics we can say that rate constant is given by;
$\lambda = \dfrac{{\ln (2)}}{T}$
Here, T is the half-life and $\lambda $ is the rate constant.
Thus, we substitute the values we get;
$\lambda = \dfrac{{\ln (2)}}{{22}}year{s^{ - 1}}$
Now as we have to find the amount at a certain time we will use the integrated rate law;
The integrated rate law is given by;
Amount left = $2{e^{\left( {\dfrac{{\ln (2)}}{T}} \right)t}}$.
Now we have to find the amount at t= 11 years and we know that T=22 years.
So by substituting the values we get;
Amount left $ = 2{e^{\left( {\dfrac{{\ln 2}}{{22}}} \right)11}}$
$ = 2{e^{\ln \sqrt 2 }}$
$ = \dfrac{2}{{\sqrt 2 }}$
$ = \sqrt 2 $
Therefore the amount left after 11 years is 1.141 grams.
So, option B is correct.
Note: No other rate laws should be used. In a decay reaction everything is similar to a first order chemical reaction. The half-life is inversely proportional to the rate constant.
Complete step by step answer:
We know that, the reaction of Pb-210 is as follows;
$^{210}Pb \to X$
Hence the kinetics can be shown as given in the table below;
| $^{20}Pb$ | $X$ | |
| T=0 | 2 grams | 0 gram |
| T=22 years | 1 gram | 1 gram |
Now as we know the half-life we can find out the rate constant
Since the decay follows first order kinetics we can say that rate constant is given by;
$\lambda = \dfrac{{\ln (2)}}{T}$
Here, T is the half-life and $\lambda $ is the rate constant.
Thus, we substitute the values we get;
$\lambda = \dfrac{{\ln (2)}}{{22}}year{s^{ - 1}}$
Now as we have to find the amount at a certain time we will use the integrated rate law;
The integrated rate law is given by;
Amount left = $2{e^{\left( {\dfrac{{\ln (2)}}{T}} \right)t}}$.
Now we have to find the amount at t= 11 years and we know that T=22 years.
So by substituting the values we get;
Amount left $ = 2{e^{\left( {\dfrac{{\ln 2}}{{22}}} \right)11}}$
$ = 2{e^{\ln \sqrt 2 }}$
$ = \dfrac{2}{{\sqrt 2 }}$
$ = \sqrt 2 $
Therefore the amount left after 11 years is 1.141 grams.
So, option B is correct.
Note: No other rate laws should be used. In a decay reaction everything is similar to a first order chemical reaction. The half-life is inversely proportional to the rate constant.
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