MCQ
$99 \%$ of a radioactive element will decay between
- ✓$6$ and $7$ half lives
- B$7$ and $8$ half lives
- C$8$ and $9$ half lives
- D$9$ half lives
$N=N_{0}\left(\frac{1}{2}\right)^{n}$
$\frac{N}{N_{0}}=\left(\frac{1}{2}\right)^{n}$
$\frac{1}{100}=\left(\frac{1}{2}\right)^{n}$
$2^{n}=100$
The number of half lives is calculated as,
$2^{n}=100$
$n$ lies between the $6$ and $7 .$ So, the radioactive element will decay between $6$ and $7$ half lives.
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(Rydberg constant $R = 1.097 \times {10^7}$per metre)

