Question

The half-life of thorium $\left(Th^{232}\right)$ is $1.4 \times 10^{10}$ years then the fraction of thorium atoms decaying per year is very nearly-
(A) $1 \times 10^{-10}$
(B) $4.95 \times 10^{-11}$
(C) $0.69 \times 10^{-11}$
(D) $7.14 \times 10^{-11}$

Answer 1

Hint: The rate of radioactivity of a sample depends on the number of atoms in the sample available for decay. The more the number of atoms in the sample the faster is the decay of the sample.
Formula used:
The radioactive decay results in decrease in the number of radioactive nuclei in the same given as:
$N = {N_0 }e^{- \lambda t}$

Complete step-by-step solution:
In the radioactive decay, we have the following differential equation:
$\dfrac{dN}{dt} = - \lambda N$
Upon integration, this differential equation gives:
$N = {N_0} e^{- \lambda t}$
We can obtain the constant as:
$\dfrac{1}{t} \ln \left( \dfrac{N_0}{N} \right) = \lambda $ .
Keeping the given values in this, we get:
$\dfrac{1}{\ln 2 \times 1.4 \times 10^{10} year} = \lambda $
In the given question, we are asked to find the fraction of thorium atoms decaying per year.
We substitute time duration as 1 year. We keep the value of the decay constant back in the equation to obtain:
$\dfrac{1}{t} \ln{\dfrac{N_0}{N} } = \lambda $
This happens to be the ratio of the number of fractions of atoms per year.
$\implies \dfrac{1}{t} \ln{\dfrac{N_0}{N} } = 1.03 \times {10^{ - 10}}$
Therefore the correct answer is option (A).

Note: The fraction of thorium atoms decaying per year is nothing but the decay constant. Whenever we write an exponential, the quantity that goes in the exponential is dimensionless. Therefore, the decay constant times the time here is written in the exponential. The decay constant is a number that has dimensions of inverse of time.