Question

If 10% of a radioactive material decays in 5 days, then the amount of the original material left after 20 days is approximately:
A. 60%
B. 65%
C. 70%
D. 75%

Answer 1

Hint: Write down radioactive formula and then find the data given from the question given and put that data into the formula and calculate the amount required. Learn the concept, how to apply logarithm and exponential terms. Learn the difference between log and ln.

Complete step by step answer:
The radioactive formula is given by
$N={{N}_{o}}{{e}^{-\lambda t}}$
Where, ${{N}_{o}}$ = the initial quantity of the substance
$N$ = quantity still remained and not yet decayed
t = time
$\lambda $= decay constant
Given data
Decayed quantity = 10% of initial quantity of the substance
Decayed quantity = $0.1{{N}_{o}}$
 $N$= ${{N}_{o}}$- Decayed quantity
$\begin{align}
  & N={{N}_{o}}-0.1{{N}_{o}} \\
 & N=0.9{{N}_{o}} \\
\end{align}$
 Applying radioactive formula
$\begin{align}
  & N={{N}_{o}}{{e}^{-\lambda t}} \\
 & 0.9{{N}_{o}}={{N}_{o}}{{e}^{-\lambda t}} \\
 & 0.9={{e}^{-\lambda t}} \\
\end{align}$
Time taken for decay 10% = t = 5 days
$\begin{align}
  & 0.9={{e}^{-5\lambda }} \\
 & \ln 0.9=-5\lambda \\
 & \lambda =-\dfrac{\ln 0.9}{5} \\
\end{align}$
When t = 20 days
Let x is percent of the quantity still remained and not yet decayed
Applying radioactive formula
$\begin{align}
  & N={{N}_{o}}{{e}^{-\lambda t}} \\
 & x{{N}_{o}}={{N}_{o}}{{e}^{-20\lambda }} \\
 & x={{e}^{-20\lambda }} \\
 & \ln x=-20\lambda \\
 & \ln x=-20(-\dfrac{\ln 0.9}{5}) \\
 & (as\lambda =-\dfrac{\ln 0.9}{5}) \\
 & \ln x=4\ln 0.9 \\
 & \ln x=\ln {{0.9}^{4}} \\
 & x={{0.9}^{4}} \\
 & x=0.658 \\
\end{align}$
Hence option B is correct.

Note: Here always use ln not log because $\ln {{e}^{x}}=x$ and \[\log {{e}^{x}}\ne x\]. Also remember that N is the remaining quantity not decayed quantity. Students can get confused. Decayed quantity =${{N}_{o}}$- Decayed quantity.