Question

A radioactive substance contains $10000$ nuclei and its half-life period is $20$ days. The number of nuclei present at the end of $10$ days is

• A. $7070$
• B. $9000$
• C. $8000$
• D. $7500$

Answer 1

The correct option is A $7070$

NN0=(12)t/T⇒N10000=(12)1020We know,
$\frac{N}{N_0}={\left(\frac{1}{2}\right)}^{t/T}\Rightarrow \frac{N}{10000}={\left(\frac{1}{2}\right)}^{\frac{10}{20}}$
$\Rightarrow N=\frac{10000}{\sqrt{2}}=\frac{10000}{1.414}=7070$