Question

The half-life period of $RaB{(}_{82}P{b}^{214})$ is $26.8min$. The mass of one curie of $RaB$ is

• A. $3.71\times {10}^{10}g$
• B. $3.71\times {10}^{-10}g$
• C. $8.61\times {10}^{10}g$
• D. $3.064\times {10}^{-8}g$

Answer 1

Answer: (D)

$3.064\times {10}^{-8}g$

$1Ci=5.3183\times {10}^{-12}moles\times t_{1/2}inminutes$
Given, half-life of $RaB$ is $26.8min$
Therefore, one Curie of $RaB$ is $5.3183\times {10}^{-12}\times 26.8moles$$=1.43\times {10}^{-10}moles$
$1moleofPb$ is $214g$.
Therefore, $1Ci=1.43\times {10}^{-10}\times 214=3.06\times {10}^{-8}g$