Question

Calculate the mass of ${{C}^{14}}$ (half-life = 5720 years) atoms which give $3.7\times {{10}^{7}}$ disintegrations per second.

Answer 1

We know that,${{t}_{\dfrac{1}{2}}} =\frac{0.693}{\lambda }$
Where ${{t}_{\dfrac{1}{2}}}$ is the half-life period and $\lambda$ is the disintegration constant
The rate of disintegration,$\dfrac{-dN}{dt} =\lambda N$

Complete step by step answer:
In the question we are provided with the information of the half-life of ${{C}^{14}}$ and also the rate of disintegration of ${{C}^{14}}$ and we have to find the mass of the ${{C}^{14}}$ that is undergoing disintegration at the given rate.
- Half-life is the time period required for the substance to reduce to its half the amount of the initial concentration.
Now let’s solve the given problem.
Let’s assume the mass of ${{C}^{14}}$ atoms be m grams.
- Now we should know the, number of atoms present in m grams of ${{C}^{14}}$,
So we use the formulae,
$\text{Number of atoms=}\dfrac{\text{Mass of the substance}}{\text{Molecular}\,\text{mass}}\text{ }\\!\\!\times\\!\\!\text{ }{{\text{N}}_{\text{A}}}$
${{N}_{A}}=6.022\times {{10}^{23}}$=Avogadro number
$Number\,of\,atoms\,in\,{{C}_{14}}=\dfrac{m}{14}\times 6.022\times {{10}^{23}}=N$
- The half-life equation is,
${{t}_{\dfrac{1}{2}}}=\dfrac{0.693}{\lambda }$
The ${{t}_{\dfrac{1}{2}}}$ is the half-life period and $\lambda$ is the disintegration constant
We alter the equation and write for disintegration constant, $\lambda$
$\lambda =\dfrac{0.693}{{{t}_{\dfrac{1}{2}}}}$
- The given ${{t}_{\dfrac{1}{2}}}$ = 5720 years and we have to convert the given ${{t}_{\dfrac{1}{2}}}$value to seconds and substitute the values in the equation,
$\lambda =\dfrac{0.693}{5720\times 365\times 24\times 60\times 60}$
$\lambda = 3.84\times {{10}^{-12}}\sec$

We know that the, $\dfrac{-dN}{dt}=\lambda N$
N = number of atoms
$\dfrac{-dN}{dt}$ is the disintegration constant
$\dfrac{-dN}{dt}$ = $3.7\times {{10}^{7}}$sec
$\lambda =3.84\times {{10}^{-12}}\sec$
$Number\,of\,atoms\,in\,{{C}_{14}}=\dfrac{m}{14}\times 6.022\times {{10}^{23}}=N$
Substitute all these values in the equation of, rate of disintegration,
$3.7\times {{10}^{7}}=\dfrac{3.84\times {{10}^{-12}}\times 6.022\times {{10}^{23}}\times m}{14}$
$3.7\times {{10}^{7}}\times 14=3.84\times {{10}^{-12}}\times 6.022\times {{10}^{23}}\times m$
$m=\dfrac{3.7\times {{10}^{7}}\times 14}{3.84\times {{10}^{-12}}\times 6.022\times {{10}^{23}}}g$
$m=2.2636\times {{10}^{-4}}g$

Note: We should convert the ${{t}_{\dfrac{1}{2}}}$into seconds since the disintegration constant is given in seconds and the ${{t}_{\dfrac{1}{2}}}$ is given in seconds.
- For avoiding all these steps, the problem may be solved by one step, if we know the sub-equations of all the terms,
$\dfrac{-dN}{dt}=\lambda N=\dfrac{0.693}{{{t}_{\dfrac{1}{2}}}}\times \dfrac{W}{M}\times {{N}_{A}}$
Where W is the weight or mass of the substance is the molecular mass.