The half-value period of $\mathrm{RaB} \left({}_{82}\mathrm{Pb}^{214}\right)$ is 26.8 minutes. The mass of one curie of RaB is

Here T = 26.8 minutes

= 26.8 x 60 sec

$\therefore$  Decay constant,

$$\begin{gathered} \mathrm{\lambda }=\frac{0.693}{\mathrm{T}}=\frac{0.693}{26.8 \mathrm{x} 60} \\ =4.32 \mathrm{x} {10}^4{\sec }^{-1} \end{gathered}$$

Now 1 curie is equal to $3.71 \mathrm{x} {10}^{10}$ disintegrations per second.

$\therefore \mathrm{\lambda }=3.71 \mathrm{x} {10}^{10}$

If N be the number of atoms in one curie, then 

$-\frac{\mathrm{dN}}{\mathrm{dt}}=\mathrm{\lambda N}$

$\mathrm{or} 3.71 \mathrm{x} {10}^{10}=4.31 \mathrm{x} {10}^{-4}$ N

$\therefore \mathrm{N}=\frac{3.71 \mathrm{x} {10}^{10}}{4.31 \mathrm{x} {10}^{-4}}=8.607 \mathrm{x} {10}^{13}$

Further, atomic weight of RaB = 214 and Avogadro's number = $6.025 \mathrm{x} {10}^{23}$

Mass of one atom $=\frac{214}{6.025 \mathrm{x} {10}^{23}}$

Mass of N atoms 

$$\begin{gathered} =\left(\frac{214}{6.025 \mathrm{x} {10}^{23}}\right) \mathrm{x} \left(8.607 \mathrm{x} {10}^{13}\right) \\ =3.064 \mathrm{x} {10}^{-8}\mathrm{g} \end{gathered}$$