Solveeit Logo

Question

Question: The half-life of a radioactive substance is 3 second. Initially it has 8000 atoms. Find (i) its deca...

The half-life of a radioactive substance is 3 second. Initially it has 8000 atoms. Find (i) its decay constant and (ii) time after which 1000 atoms will remain undecayed.

Explanation

Solution

We can calculate the decay constant directly with the expression because time 3 seconds is given to us in the question. The required time after which 1000 atoms will remain undecayed, that can be determined by using a logarithm in the formula of the law of radioactive decay.

Complete step by step answer:
(i) It is given to us that the half-life of the radioactive decay is three seconds, so we use it directly in the expression of radioactive decay. So, we will write the expression of the decay constant.
λ=0.6931T\lambda = \dfrac{{0.6931}}{T}
Here, λ\lambda is the decay constant and TT is the half-life of a radioactive substance.
We will substitute the value of half-life in the above expression, therefore, we get,
λ=0.69313  s λ=0.231  s1\begin{array}{l} \lambda = \dfrac{{0.6931}}{{3\;{\rm{s}}}}\\\ \lambda = 0.231\;{{\rm{s}}^{ - 1}} \end{array}
Therefore, the decay constant is 0.231  s10.231\;{{\rm{s}}^{ - 1}}.
(ii) We know that the formula of the law of radioactive decay is N=No(12)tTN = {N_o}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}}, so that we will use this formula to calculate the required time therefore we get,
N=No(12)tTN = {N_o}{\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}}
Here, NN is the number of atoms remaining aftertime t, No{N_o} total number of atoms in the initial condition, TT is the half life time and tt is the required time.
To obtain the expression of the required time form the above equation, we will use the algorithm in the above formula, so
ln(NNo)=ln2×tT t=ln(NoN)Tln2\begin{array}{c} \ln \left( {\dfrac{N}{{{N_o}}}} \right) = \ln 2 \times \dfrac{t}{T}\\\ t = \dfrac{{\ln \left( {\dfrac{{{N_o}}}{N}} \right)T}}{{\ln 2}} \end{array}
On substituting the values in the above equation, therefore we get,
t=ln(80001000)×3  sln2 t=6.238  s0.693 t=9  s\begin{array}{l} t = \dfrac{{\ln \left( {\dfrac{{8000}}{{1000}}} \right) \times 3\;{\rm{s}}}}{{\ln 2}}\\\ t = \dfrac{{6.238\;{\rm{s}}}}{{0.693}}\\\ t = 9\;{\rm{s}} \end{array}
Therefore, the required time after which 1000 will remain undecayed is 9 s.

Note: Remember the expression of the decay constant and the formula of the law of radioactive decay. The law of radioactive decays gives three important and equivalent formulas of exponential decay, so remember each and every formula related to radioactive decay and use each one according to question.