Question

The half life of $Th^{232}$ is $1.4 \times 10^{10}$ years and that of its daughter element $Ra_{238}$ is 7 years. What amount (most nearly) weight of $Ra_{238}$ will be in equilibrium with 1gm of $Th_{232}$ ?

• A. $5 \times 10^{10} gm $
• B. $5.0 \, gm $
• C. $1.95 \times 10^{-9} gm $
• D. $2 \times 10^{- 10} gm $

Answer 1

Answer: (A)

$\because \frac{t_{1 / 2}( Th -232)}{t_{1 / 2}( Ra -238)}=\frac{1 \times N_{0} / 232}{W \times N_{0} / 238}$

$\because \, \frac{W}{M}=\frac{N}{N_{0}}$ or $N=\frac{W \times N_{0}}{M}$

Where,

$W=$ mass of substance taken

$M=$ molar mass

$N=$ actual number of particles

$N_{0}=$ Avogadro number.

Also, $ M( Th -232), M( Ra -238)$

$\therefore \frac{t_{1 / 2}( Th )}{t_{1 / 2}( Ra )}=\frac{1.4 \times 10^{10}}{7}=\frac{1 \times \frac{N_{0}}{232}}{W \times \frac{N_{0}}{238}}$

or $W=\frac{238}{232 \times 2 \times 10^{9}}$

$=5 \times 10^{-10}\, g$