Question

A radio active substance contains $10,000$ nuclei and its half life period is $20$ days. The number of nuclei present at the end of $10$ days is

• A. $7,500$
  
• B. $8,000$
  
• C. $9,000$
  
• D. $7,070$
  

Answer 1

The correct option is D $7,070$

number of nuclei initially $N_o=10000$

Half life $T_{1/2}=20$ days

Time of observation $t=10$ days

Thus number of nuclei after $t=10$ days $N=N_o{e}^{\frac{-\left(\ln 2\right)t}{T_{1/2}}}$

Or $N=10000\times e^{\frac{-\left(\ln 2\right)\times 10}{20}}$

Or $N=10000\times e^{\frac{-\ln 2}{2}}$

Or $N=10000\times e^{-\ln \sqrt{2}}$ (using $x\ln y=\ln y^x$)

Or $N=10000\times e^{\ln \frac{1}{\sqrt{2}}}$

Or $N=10000\times \frac{1}{\sqrt{2}}$

$⟹$ $N=10000\times 0.707=7070$ Nuclei