Question

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

• A. $5\times 10^{10}gm$
• B. $5.0gm$
• 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 $W=\frac{238}{232\times 2\times 10^9}$

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