All the disintegrations of a sample of an unknown nuclide weighing 4.6 *10¯² g were counted. In the first half-life of the sample, the total number of disintegrations counted was 4.3 *10²⁰. What is the atomic weight of the unknown element?

Question

Quizplus · Accepted Answer

The number of atoms in the sample is:$$N_0 = \frac{4.6 	imes 10^{-2} g}{A} 	imes 6.022 	imes 10^{23} atoms/mol$$where A is the atomic weight of the unknown element.The number of disintegrations in the first half-life is:$$N_1 = N_0 (1 - e^{-0.693}) = 4.3 	imes 10^{20}$$Substituting these values into the equation above, we get:$$\frac{4.6 	imes 10^{-2} g}{A} 	imes 6.022 	imes 10^{23} atoms/mol (1 - e^{-0.693}) = 4.3 	imes 10^{20}$$Solving for A, we get:$$A = 32 amu$$

All the Disintegrations of a Sample of an Unknown Nuclide