$$\eqalign{ & {t_{\frac{1}{2}}} = 12.3\,yr \cr & {\text{Initial amount}}\left( {{N_0}} \right) = 32\,mg \cr & {\text{Amount left}}\left( N \right) = ? \cr & {\text{Total time}}\left( T \right) = 49.2\,yr \cr & \frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^n} \cr & {\text{where, }}n = {\text{total number of half - lives}} \cr & n = \frac{{{\text{Total time}}}}{{{\text{Half - life}}}} \cr & \,\,\,\,\, = \frac{{49.2}}{{12.3}} \cr & \,\,\,\,\, = 4 \cr & {\text{So,}}\,\,\,\frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^n} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\frac{N}{{32}} = {\left( {\frac{1}{2}} \right)^4} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\frac{N}{{32}} = \frac{1}{{16}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,N = \frac{{32}}{{16}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\,mg \cr} $$

The atomic number of isotope of hydrogen is always one, so it have only one proton in its nucleus and given that it have two neutrons, it means that a its mass number is 3.

The age of most ancient geological objects is determined by uranium-lead method. This method is based on the radioactive properties of uranium. In this method, we assume that initially the rock did not contain any lead, we can calculate the age of the rock by measuring the ratio of concentrations of $$_{92}{U^{238}}$$  and $$_{82}P{b^{206}}.$$

$$_{28}N{i^{60}} + \mathop {_0{n^1}}\limits_{\left( n \right)} { \to _{27}}C{o^{60}} + \mathop {_1{H^1}}\limits_{\left( p \right)} $$
( $$n, p$$  means that neutron attacks and proton liberates )

When an alpha particle is emitted from a radioactive element, then its atomic number is decreased by 2 units and mass number is decreased by 4 units.
$$_{84}P{o^{210}} \to {\,_{82}}P{b^{206}} + {\,_2}H{e^4}$$

$$235\,g\,\,{\text{of}}\,\,U - 235\,\,{\text{contains}}$$      $${\text{6}}{\text{.023}} \times {\text{1}}{{\text{0}}^{23}}\,{\text{atoms}}$$
$$\eqalign{ & 1\,g\,U - 235 = \frac{{6.023 \times {{10}^{23}}}}{{235}}\,atoms \cr & \therefore \,\,{\text{Energy released}} \cr & = \frac{{3.2 \times {{10}^{ - 11}} \times 6.023 \times {{10}^{23}}}}{{235}}J \cr & = 8.21 \times {10^{10}}J \cr & = 8.21 \times {10^7}\,kJ \cr} $$

$$\eqalign{ & {t_{\frac{1}{2}}}\,o{\text{f}}{\,_{11}}N{a^{24}} = 15\,h \cr & {\left[ R \right]_0} = 1.0\,\mu g \cr & \left[ R \right] = 1.0 \times \frac{{25}}{{100}} = \frac{{25}}{{100}} = 0.25\,\mu g \cr & k = \frac{{0.6932}}{{{t_{\frac{1}{2}}}}} = \frac{{0.6932}}{{15}}\,{h^{ - 1}} \cr & \therefore \,\,k = \frac{{2.303}}{t}{\text{log}}\frac{{{{\left[ R \right]}_0}}}{{\left[ R \right]}} \cr & \therefore \,\,\frac{{0.6932}}{{15}} = \frac{{2.303}}{t}{\text{log}}\frac{1}{{0.25}} \cr & \frac{{0.6932}}{{15}} = \frac{{2.303}}{t}{\text{log}}\frac{{100}}{{25}} = \frac{{2.303}}{t}{\text{log}}4 \cr & \frac{{0.6932}}{{15}} = \frac{{2.303}}{t} \times 0.6020 \cr & t = \frac{{15 \times 2.303 \times 0.6020}}{{0.6932}} \cr & \,\,\, = \frac{{20.7995}}{{0.6932}} \cr & \,\,\, = 30.00\,h \cr} $$

$$_{92}{U^{235}}$$  nucleus absorbs a neutron and then disintegrate in $$_{54}X{e^{139}},{\,_{38}}S{r^{94}}$$    and $$X.$$
Thus,
$$_{92}{U^{235}}{ + _0}{n^1}{ \to _{54}}X{e^{139}}{ + _{38}}S{r^{94}} + {3_0}{n^1}$$
Hence, the product is 3-neutrons.

$${\text{Remaining activity}} = $$     $$0.01\,M\,\,{\text{after 24 h}}$$
$${\text{Remaining activity}} = $$     $${\text{Initial activity}} \times {\left( {\frac{1}{2}} \right)^n}$$
$$\eqalign{ & {\text{Used half - life time}} \cr & \left( n \right) = \frac{{{\text{Total time}}}}{{{t_{\frac{1}{2}}}}} \cr & \,\,\,\,\,\,\,\,\, = \frac{{24}}{6} \cr & \,\,\,\,\,\,\,\,\, = 4 \cr & {\text{So,}}\,\,\,0.01 = {\text{Initial activity}} \times {\left( {\frac{1}{2}} \right)^4} \cr & {\text{Initial activity}} = 0.01 \times 16 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.16 \cr} $$