Question

The half−value period of a synthetic molecule is given to be 26.8 minutes. The mass of One curie of this molecule is (1 curie $=3.71\times {10}^{10}dps$ and molecular mass of molecule $=214\text{g}$)

• A. $3.71\times {10}^{10}g$
• B. $3.71\times {10}^{10}g$
• C. $8.61\times {10}^{10}g$
• D. $3.064\times {10}^{-8}g$

Answer 1

The correct option is D $3.064\times {10}^{-8}g$

$t_{\frac{1}{2}}=26.8\times 60=1608s$
$1608=\ln 2\tau$
$1608=\ln 2\frac{1}{\lambda }$
$\lambda =\frac{\ln 2}{1608}$
$1\text{curie}=\frac{\ln 2}{1608}{N}_0$
$3.7\times {10}^{1}0=\frac{\ln 2}{1608}{N}_0$

$\because N_0=\frac{214}{6.022\times {10}^{23}}\times mg$

Putting all the values we get mass of one curie of the molecule as $3.064\times {10}^{-8}g$.