Question

The mass of $U^{234}$, with half-life $2.5\times {10}^5$, corresponding to its activity of $1$ Curie will be

• A. $163.62$ gm
• B. $1.438\times {10}^{-11}$ gm
• C. $3.7\times {10}^{-10}$ gm
• D. $3.7\times {10}^{10}$ gm

Answer 1

The correct option is A $163.62$ gm

Given : Half life of $U^{234}$ $T_{1/2}=2.5\times {10}^5$ yr $=2.5\times {10}^5\times 365\times 86400=7.884\times {10}^{12}$ s

Activity $A=1Ci=3.7\times {10}^{10}$ dps

using $A=\frac{0.693}{T_{1/2}}N$

$3.7\times {10}^{10}=\frac{0.693}{7.884\times {10}^{12}}N$ $⟹N=4.21\times {10}^{23}$ $U^{234}$ atoms

Mass of $U^{234}$, $m=\frac{N}{N_A}M=\frac{4.21\times {10}^{23}}{6.023\times {10}^{23}}\times 234$ gm $=163.62$ gm