Question

${}^{90}\text{Y}$ has a half-life of 64 hour and ${}^{90}\text{Sr}$ has a half-life of 28 years. ${}^{90}\text{Sr}$ decays to ${}^{90}\text{Y}$ by $\beta$-emission. The mass of ${}^{90}\text{Y}$ in equilibrium with 1 g of ${}^{90}\text{Sr}$ after 1 hour is :

• A. $[5.22\times {10}^{-4}g]$
• B. $[2.81\times {10}^{-4}g]$
• C. $[4.31\times {10}^{-4}g]$
• D. none of these

Answer 1

Answer: (B)

$[2.81\times {10}^{-4}g]$

${}^{90}Sr=A$ and ${}^{90}Y=B$

The sequential decay equation is

$A\to B\to C$

$B=\frac{A_0{k}_1}{k_2-k_1}(e^{-k_{1}t}-e^{-k_{2}t})$......(1)

$k_1=\frac{0.693}{t_{1/2}}=\frac{0.693}{28\times 365\times 24}$

$k_2=\frac{0.693}{t_{1/2}}=\frac{0.693}{64}$

Substitute in equation (1),

$B=\frac{1\times \frac{0.693}{28\times 365\times 24}}{\left(\frac{0.693}{64}\right)-\left(\frac{0.693}{28\times 365\times 24}\right)}(e^{-\left(\frac{0.693}{28\times 365\times 24}\right)t}-e^{-\left(\frac{0.693}{64}\right)t})=2.81\times {10}^{-4}g$