Question

After the activity of ${}^{56}Mn$ becomes constant, number of ${}^{56}Mn$ nuclei present in the target is equal to

• A. $5\times {10}^{11}$
• B. $20\times {10}^{11}$
• C. $1.2\times {10}^{14}$
• D. $1.8\times {10}^{15}$

Answer 1

The correct option is D $1.8\times {10}^{15}$

$Activity\left(A\right)=\frac{ln2}{halflife}\times Numberofparticles\left(N\right)$

Here, $A=P=13.86\times {10}^{10}/sec$, $halflife=2.5h=9000s$

Thus $N=13.86\times {10}^{10}\times 9000/0.693=1.8\times {10}^{15}$