Question

The curie was originally intended to be the activity of one gram of radium-226. The actual weight of radium which is equal to one curie is (in gm)
$T_{50}$ (Radius-226) = 1620 years
One curie = $3.7\times {10}^{10}$ disintegrations (of atoms) per second (dps)
One year = $3.154\times {10}^7$ seconds

Answer 1

$T_{10}$ (Radium - 226)\, 1620 years = $1620\times 3.154\times {10}^{7}s$
Hence k (disintegration constant) = $\frac{0.693}{T_{50}}$
$=\frac{0.693}{1620\times 3.154\times {10}^7}{s}^{-1}$
Also $\left(\frac{dx}{dt}\right)=3.7\times {10}^{10}dps=k\frac{w{N}_0}{A}$
$\therefore 3.7\times {10}^{10}dps=\frac{0.693\times w\times 6.02\times {10}^{23}}{1620\times 3.154\times {10}^7\times 226}$
where w is the weight of Ra-226 to have activity of one curie and A its atomic weight
$\therefore w=1.024g$