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Question: The half-life of \( {C^{14}} \) is \( 5670 \) years. If \( 56g \) of \( {C^{14}} \) was present init...

The half-life of C14{C^{14}} is 56705670 years. If 56g56g of C14{C^{14}} was present initially, how many atoms of C14{C^{14}} is left after 2268022680 years.
\left( A \right)1.5 \times {10^{23}} \\\ \left( B \right)2.9 \times {10^{27}} \\\ \left( C \right)1.5 \times {10^{27}} \\\ \left( D \right)2.9 \times {10^{23}} \\\

Explanation

Solution

Hint : In order to solve this question, we are going to first determine the decay constant by the reciprocal of decay constant. Then, by using the law of radioactive decay, the number of disintegrated atoms present at a particular time is calculated while the number of atoms initially can be calculated from initially given.
The decay constant for the radioactive element is given by:
λ=1t12\lambda = \dfrac{1}{{{t_{\dfrac{1}{2}}}}}
According to law of radioactive decay, the rate of disintegration is given by
AA0=e−λt\dfrac{A}{{{A_0}}} = {e^{ - \lambda t}}

Complete Step By Step Answer:
Let us start solving the question by calculating the decay constant.
It is given that the half-life of C14{C^{14}} is 56705670 years, i.e.,
t12=5670{t_{\dfrac{1}{2}}} = 5670
The decay constant for the radioactive element C14{C^{14}} is given by:
λ=1t12\lambda = \dfrac{1}{{{t_{\dfrac{1}{2}}}}}
Calculating this value of the decay constant by putting the value of half life
λ=15670\lambda = \dfrac{1}{{5670}}
The rate of disintegration for C14{C^{14}} is given by
\dfrac{A}{{{A_0}}} = {e^{ - \lambda t}} \\\ \ln \left( {\dfrac{A}{{{A_0}}}} \right) = - \lambda t \\\
Now, the number of atoms of C14{C^{14}} present initially can be calculated as
{A_0} = \dfrac{{56}}{{14}} \times 6.022 \times {10^{23}} \\\ \Rightarrow {A_0} = 4 \times 6.022 \times {10^{23}} = 24.088 \times {10^{23}} \\\
Putting the value of A0{A_0} obtained above in the above equation, we get,
\ln \left( {\dfrac{A}{{{A_0}}}} \right) = - \dfrac{1}{{5670}} \times 22680 \\\ \Rightarrow \ln \left( {\dfrac{A}{{24.088 \times {{10}^{23}}}}} \right) = - 4 \\\
On further solving this equation, we get
A=1.5×1023atomsA = 1.5 \times {10^{23}}atoms .

Note :
According to the law of radioactive decay, the quantity of a radio-element which disappears in unit time is directly proportional to the amount present. Rate of decay of nuclide is independent of temperature, so its activation energy is zero. The step where the initial number of atoms is calculated is important and that depends on the initial mass and the atomic number of C14{C^{14}} .