Question

The half life period of Pb${}^{210}$ is $22$years. If $2g$ of Pb${}^{210}$ is taken, then after $11$ years the amount of

pb${}^{210}$ will be present is

• A. $0.1414g$
• B. $1.414g$
• C. $2.828g$
• D. $0.707g$

Answer 1

Answer: (B)

$1.414g$

Half life period $=22years$

$\lambda =\frac{ln2}{22}year{s}^{-1}$

So, using integrated rate law:

Amount left/present $=2e^{-\frac{ln2}{222}\times 11}$

$=2e^{-ln\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}$

$=1.414gms$