Question

The half-life of a radon is $ 3.8 $ days. Calculate how much radon of $ 15 $ milligram will remain after $ 38 $ days?

Answer 1

Hint: To solve this question, we must understand the concept of radioactive substance. So we will use the formula to solve this question, the formula is:
 $ m(n) = {\left( {\dfrac{1}{2}} \right)^n} \times {m_0} $ , where
  $ m(n) $ is the mass after a certain number of half lives.
 $ n $ is the number of half lives that have passed,
And $ {m_0} $ is the mass before any half-lives that have passed i.e. the original mass.

Complete Step By Step Answer:
We know that the half-life of a radioactive element is the time that it takes until half of the element has decayed. After half life, half of the original mass of the element will be left.
So after $ 2 $ half lives, half of the mass after $ 1 $ half life will be left, so it means that we are left with a quarter of the original mass after two half lives.
Here, we have been given the original mass of the radon is $ 15\,mg $ i.e.
  $ {m_0} = 15\,mg $
We can calculate the number of half-lives that have passed i.e.
  $ n $ .
It is given that the half-life of a radon is $ 3.8 $ days. We have to calculate after $ 38 $ days.
So it gives of number of half-lives that have passed
 $ \dfrac{{38}}{{3.8}} $
It gives us value
 $ \Rightarrow \dfrac{{38 \times 10}}{{38}} = 10 $
Now we have
 $ {m_0} = 15\,mg,n = 10 $
By substituting the values in the formula we have:
 $ \Rightarrow m(n) = {\left( {\dfrac{1}{2}} \right)^{10}} \times 15\,mg $
We will now simplify this:
 $ \Rightarrow \dfrac{{15 \times {1^{10}}}}{{{2^{10}}}} = \dfrac{{15}}{{1024}} $
Therefore it gives us value
 $ m(n) = 0.0146\,mg $
Hence the required answer is $ 0.015\,mg $ (approx.)

Note:
We should note that a half-life usually describes the decay of discrete entities, such as radioactive atoms. We should know that we can write the formula in different way too i.e. $ \dfrac{N}{{{N_0}}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{t}{T}}} $ ,
 Where half life period is
 $ T = 3.8 $ ,
Total time is
  $ t = 38 $ days,
We have initial amount i.e.
 $ {N_0} = 15\,mg $
And the amount left un-decayed is $ N $ .
By substituting the values, in the formula we have:
 $ \Rightarrow \dfrac{N}{{15}} = {\left( {\dfrac{1}{2}} \right)^{\dfrac{{38}}{{3.8}}}} $
It gives us
  $ \Rightarrow \dfrac{N}{{15}} = {\left( {\dfrac{1}{2}} \right)^{10}} $
ON cross multiplication, we have:
 $ \Rightarrow N = 15 \times \dfrac{1}{{{2^{10}}}} $
It gives us same value as above:
  $ \Rightarrow N = \dfrac{{15}}{{1024}} = 0.015\,mg $ .