Question

Uranium ores contain one radium $-226$ atoms for every $2.8\times {10}^6$ uranium $-238$ atoms. Calculate the half - life of ${}_{92}^{238}U$, given that the half - life of ${}_{88}^{226}Ra$ is 1600 years. (${}_{88}^{226}Ra$ is a decay product of ${}_{92}^{238}U$)

• A. $1.75\times {10}^3\text{years}$
• B. $1600\times \frac{238}{92}\text{years}$
• C. $4.55\times {10}^9\text{years}$
• D. $1600\text{years}$

Answer 1

The correct option is C $4.55\times {10}^9\text{years}$

Activity of uranium atom and activity of radium is equal.
$\Rightarrow {\lambda }_U{N}_U={\lambda }_{Ra}{N}_{Ra}$
or $\frac{N_U}{{\tau }_U}=\frac{N_{Ra}}{{\tau }_{Ra}}$
or $\frac{N_U}{\ln 2{\tau }_U}=\frac{N_{Ra}}{\ln 2{\tau }_{Ra}}$
$\frac{N_U}{t_U}=\frac{N_{Ra}}{t_{Ra}}$
$\frac{2.8\times {10}^6}{t_U}=\frac{1}{1600}$
$t_U=4.5\times {10}^9\text{years}$