Nuclear Chemistry MCQ Questions & Answers in Physical Chemistry | Chemistry

$$\eqalign{ & {\text{By carbon dating method}} \cr & {\text{Age of wood}} = \frac{{2.303}}{{0.693}} \times {T_{0.5}}\,\,{\text{log}}\left( {\frac{{{N_0}}}{N}} \right) \cr} $$
$${\text{where,}}\,\,\frac{{{N_0}}}{N} = $$   $$\left[ {\frac{{{\text{Ratio of}}\,\,\frac{{{C^{14}}}}{{{C^{12}}}}\,\,{\text{in living wood }}}}{{{\text{Ratio of}}\,\,\frac{{{C^{14}}}}{{{C^{12}}}}\,\,{\text{in dead wood }}}}} \right]$$
Hence, it is based upon the ratio of $${C^{14}}$$  and $${C^{12}}.$$

By mean of emission of an $$\alpha $$ - particle the atomic number is decreased by $$2$$ $$units$$  while mass number is decreased by $$4$$ $$units,$$  but in case of emission of one $$\beta $$ - particle the atomic number is increased by $$1$$ $$unit$$  while mass number is not affected.
$$_{92}{X^{232}}\frac{{ - 3\alpha \,and}}{{ - 3\beta }}{\,_{89}}{Y^{220}}$$

After emitting a $$\beta $$ - particle, atomic number is increased by one unit and atomic weight has no change. So, the atomic number of parent nucleus will be 5.
\[_{5}{{X}^{14}}{{\xrightarrow{-\beta }}_{6}}{{Y}^{14}}{{\xrightarrow{-\beta }}_{7}}{{N}^{14}}\]
Number of neutrons = mass number - number of protons
                                 = 14 - 5 = 9

After radioactive decay of element by three successive emission of $$\alpha $$ - particles, the last element (stable) $$Pb$$  is obtained which belongs to group 14 of periodic table.

When a radioactive element emits an electron, then one neutron in the nucleus change into proton, so atomic number increase by one unit.
$$_{91}P{a^{234}}{ \to _{92}}{U^{234}}{ + _{ - 1}}{e^0}$$
In nucleus the following reaction takes place
$$_0{n^1} \to {\,_1}{H^1} + {\,_{ - 1}}{e^0}$$

$$\eqalign{ & {\text{Hence,}}\,\,d = b - 5,\,c = a - 12 \cr & {\text{or}}\,\,\,\,\,\,\,\,\,\,\,\,\,a = c + 12,\,b = d + 5 \cr} $$

\[\underset{\left( Mass=34.96903\,u \right)}{\mathop{_{16}{{S}^{35}}}}\,\,\xrightarrow{-\beta }\,\underset{\left( Mass=34.96885\,u \right)}{\mathop{_{17}C{{l}^{35}}}}\,\]
$$\eqalign{ & {\text{Mass defect}} = \left( {34.96903 - 34.96885} \right)\,u \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.00018\,u \cr & {\text{Energy emitted}} = 0.00018 \times 931\,MeV \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 0.16758\,MeV \cr} $$

$$\eqalign{ & {\text{Half - life of radioactive}} \cr & {t_{\frac{1}{2}}} = 5760\,yr,\,{\left[ R \right]_0} = 200\,mg \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ R \right] = 25\,mg \cr & {\text{Rate constant}} \cr & \,\left( k \right) = \frac{{0.6932}}{{{t_{\frac{1}{2}}}}} \cr & \,\,\,\,\,\,\,\,\, = \frac{{0.6932}}{{5760}}y{r^{ - 1}} \cr & \therefore \,\,k = \frac{{2.303}}{t}{\text{log}}\frac{{{{\left[ R \right]}_0}}}{{\left[ R \right]}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {{R_0} = 200\,mg\,{\text{inital amount}}} \right) \cr & \frac{{0.6932}}{{5760}} = \frac{{2.303}}{t}{\text{log}}\frac{{200}}{{25}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {R = 25\,mg\,{\text{reduced amount}}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{2.303}}{t}{\text{log}}\,8 \cr & \frac{{0.6932}}{{5760}} = \frac{{2.303}}{t} \times 0.9030 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,t = \frac{{5760 \times 2.303 \times 0.9030}}{{0.6932}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{11978.54}}{{0.6932}} \simeq 17280\,yr \cr} $$

$$\eqalign{ & {\text{Radioactive decay is first order reaction}}{\text{.}} \cr & {\text{So,}}\,\,\,\lambda {\text{ = }}\frac{{0.693}}{{{t_{\frac{1}{2}}}}} \cr & \therefore \,\,\,{t_{\frac{1}{2}}} = \frac{{0.693}}{{2.31 \times {{10}^{ - 4}}}}yr \cr & = 0.3 \times {10^4} \cr & = 3 \times {10^3}\,yr \cr} $$

$$\eqalign{ & {\text{Given,}} \cr & {\text{Half - life}}\left( {{t_{\frac{1}{2}}}} \right) = 3\,h \cr & {\text{Initial mass}}\left( {{N_0}} \right) = 300\,g \cr & {\text{Total time}}\,\left( T \right) = 18\,h \cr & {\text{Mass left}}\,\left( N \right) = ? \cr & {\text{We know that,}} \cr & \frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^n} \cr & {\text{where, }}n = {\text{number of half - lives}} \cr & n = \frac{{{\text{Total}}\,{\text{time}}}}{{{\text{Half - life}}}} \cr & \,\,\,\,\, = \frac{{18}}{3} \cr & \,\,\,\,\, = 6 \cr & {\text{So,}}\,\,\frac{N}{{300}} = {\left( {\frac{1}{2}} \right)^6} \cr & {\text{or}}\,\,\,\frac{N}{{300}} = \frac{1}{{64}} \cr & N = \frac{{300}}{{64}} \cr & \,\,\,\,\,\,\, = 4.68\,g \cr} $$