Question

If 10% of a radioactive material decays in 5 days. The percentage of original material left after 20 days
A.90%
B.80%
C.65.6%
D.50%

Answer 1

Hint: When the radioactive material continuously decayes for five days such that we get the decayed quantity. We need to use the radioactive decay law, we collect the given data from the question and substitute the data in the formula then we can calculate the amount required.

Complete step by step solution:
From the given data:
Decayed material quantity=10% of initial quantity of the substance
 Decayed material quantity=0.1${N_0}$
$\
  N = {N_0} - decayed quantity \\
\implies N = {N_0} - 0.1{N_0} \\
\implies N = 0.9{N_0} \\
\ $
 By using the formula of radioactive decay law
$N = {N_0}{e^{ - \lambda t}}$
$\
  0.9{N_0} = {N_0}{e^{ - \lambda \times 5}} \\
\implies {e^{ - \lambda \times 5}} = 0.9 \\
\implies 5\lambda = {\log _e}\dfrac{1}{{0.9}}.......(1) \\
\ $
At t=20days
Let us consider X is the quantity of material still not yet decayed
Again by using the radioactive decay law
$\
  x{N_0} = {N_0}{e^{ - \lambda \times 20}} \\
\implies {e^{ - \lambda \times 20}} = x \\
\implies 20\lambda = {\log _e}\left( {\dfrac{1}{x}} \right).......(2) \\
\ $
 Dividing 1&2
$\
\implies\dfrac{1}{4} = \dfrac{{{{\log }_e}\left( {\dfrac{1}{{0.9}}} \right)}}{{{{\log }_e}\left( {\dfrac{1}{x}} \right)}} \\
\implies\dfrac{1}{4} = \dfrac{{{{\log }_{10}}\left( {\dfrac{1}{{0.9}}} \right)}}{{{{\log }_{10}}\left( {\dfrac{1}{x}} \right)}} \\
\implies \dfrac{1}{4} = \dfrac{{{{\log }_{10}}(0.9)}}{{{{\log }_{10}}x}} \\
\ $
$\
  {\log _{10}}x = 4{\log _{10}}0.9 \\
\implies x = 0.658 \\
\ $
65.6% original matter will be left after 20 days.

So, the correct answer is “Option C”.

Note:
It’s important to understand the concept of applying logarithm and exponential terms. Also we should know the difference between the ln and log.
Students should not get confused in this concept i.e. remember that V is the remaining material quantity but not decayed quantity.