Question

The half-life of a radioactive nuclide is 20h. It is found that the fraction \[\left( {\dfrac{1}{x}} \right)\] of original activity remains after 40 hours? What is the value of x?

Answer 1

Hint: We will use the formula for radioactive decay and calculate the ratio of remaining activity to original activity.
\[N(t) = \dfrac{{{N_o}}}{{{2^n}}}\]
Where,
\[N(t)\] = Remaining Activity at time =t seconds
\[{N_o}\] = Initial Activity at \[t = 0\] s
n = Total number of half-lives.
Then, we will compare our value with that given in the question to calculate the value of x.

Complete step by step solution
We will use the formula for radioactive decay: -
 \[N(t) = \dfrac{{{N_o}}}{{{2^n}}}\] . . . (1)
Where,
 \[N(t)\] = Remaining Activity at time =t seconds
 \[{N_o}\] = Initial Activity at \[t = 0\] s
n = Total number of half-lives.
 \[n = \dfrac{{total\,time}}{{time\,for\,one\,half - life}}\]
We will now put the values as given in the question into the above formula: -
 \[n = \dfrac{{40hours}}{{20hours}}\]
 \[n = 2\] . . . (2)
The question says that the ratio of remaining activity to the original activity is \[\left( {\dfrac{1}{x}} \right)\] .
Thus,
  \[\dfrac{{N(t)}}{{{N_o}}} = \dfrac{1}{x}\]
 \[N(t) = \dfrac{{{N_o}}}{x}\]
Comparing it with equation (1) we can write: -
 \[{2^n} = x\]
Substituting the value of n from eq-(2)
 \[{2^2} = x\]
 \[ \Rightarrow x = 4\]

Hence, the value of x is equal to 4.

Note
There is an alternate formula for radioactive decay that can be used to do this question.
\[N = {N_\circ }{e^{ - \lambda t}}\]
Where,
N = remaining activity
 \[{N_\circ }\] = Original activity
 \[\lambda = \] decay constant
E = euler’s constant = 2.74
t = total time
Also, time for half life \[ = \dfrac{{\ln 2}}{\lambda } = \dfrac{{0.693}}{\lambda }\]
Now, using the above equation we can calculate \[\lambda \] .
 \[\lambda \] = 0.693 /time for half-life
 \[\lambda = \dfrac{{0.693}}{{20}}\] (time for half-life= 20h)
\[\lambda \]= 0.03465
Now,
  \[N = {N_\circ }{e^{ - 0.03465 \times 40}}\] (for time = 40h)
  \[\dfrac{N}{{{N_\circ }}} = {e^{ - 1.386}}\]
 \[\dfrac{N}{{{N_\circ }}} = \dfrac{1}{{{e^{1.386}}}} = \dfrac{1}{{4.043}}\]
Comparing this with, \[\dfrac{{N(t)}}{{{N_o}}} = \dfrac{1}{x}\] , we get: -
 \[ \Rightarrow x = 4.043\]
This method gives an approximate value and also involves much greater calculation. It gives approximate value because the values for ‘e’ and ‘ln2’ are not completely accurate. That is why it is recommended to not use this method for objective questions whereas, the solution is completely correct.