Question

A chemist prepares 1.00 g of pure ${}_6^{11}C$. This isotope has half-life of 21 min, decaying by equation:
${}_6^{11}C{⟶}_5^{11}B{+}_1^{0}e$
What is rate of disintegration per second (dps) at start?

• A. $3\times {10}^{19}$ dps
• B. $4\times {10}^{19}$ dps
• C. $6\times {10}^{19}$ dps
• D. None of these

Answer 1

The correct option is A $3\times {10}^{19}$ dps

$-\frac{dN}{dt}=\lambda N_0$
where $\lambda$(disintegration constant) $=\frac{0.693}{T_{50}}$ and $N_0$ the number of atoms present initially
$N_0=\frac{weight}{atomicweight}\times Avogadr{o}^{\prime }snumber$
$-\frac{dN}{dt}=\frac{0.693}{21\times 60}\times \frac{1}{11}\times 6.02\times {10}^{23}$ dps
$=3\times {10}^{19}$ dps