Question

Atomic weight of a radioactive element is ${M_w}gram$ Radioactivity of m gram. of its mass is:-
(${N_A} = Avogadro\;number$, $\lambda = decay\;constant$)

A.${N_A}\lambda $
B.$(\dfrac{{{N_A}}}{{{M_w}}}m)\lambda $
C.$(\dfrac{{{N_A}}}{m})\lambda $
D.$(\dfrac{{{N_A}}}{m}{M_w})\lambda $

Answer 1

Hint: Here atomic mass and total mass of the element is given find the total number of atoms by finding the number of moles and then multiplying it with avogadro number and then use the law of radioactivity to find its radioactivity.

Complete step by step answer:
Given atomic mass is ${M_w} gram$ and mass of the radioactive material is m gram so we can find no of moles by using $number \;of\;moles = \dfrac{{given\;mass}}{{atomic\;mass}}$
therefore, $no\;of\;moles = \dfrac{m}{{{M_w}}}$
now we need to find total no of atoms
we know that 1 mole contains ${N_A}$atoms
so $\dfrac{m}{{{M_w}}}$moles will have ($\dfrac{m}{{{M_w}}}$${N_A}$)atoms.
Now,
We need to use the law of radioactivity to find the radioactivity of the given element.
We know that ,
$Radioactivity = \dfrac{{dN}}{{dt}} = - \lambda N$ where, N is number of atoms
So here, $radioactivity = - \lambda (\dfrac{m}{{{M_w}}}{N_A})$

Hence, option B is correct.

Note: These types of questions are relatively easy all you need to do is use the fundamental law of radioactivity and put the values here for example the trickiest part is finding the number of atoms, once you know the number of atoms all you need to do put the number of atoms in the formula of radioactivity.