$$\eqalign{
& {\text{By carbon dating method}} \cr
& {\text{Age of wood}} = \frac{{2.303}}{{0.693}} \times {T_{0.5}}\,\,{\text{log}}\left( {\frac{{{N_0}}}{N}} \right) \cr} $$
$${\text{where,}}\,\,\frac{{{N_0}}}{N} = $$ $$\left[ {\frac{{{\text{Ratio of}}\,\,\frac{{{C^{14}}}}{{{C^{12}}}}\,\,{\text{in living wood }}}}{{{\text{Ratio of}}\,\,\frac{{{C^{14}}}}{{{C^{12}}}}\,\,{\text{in dead wood }}}}} \right]$$
Hence, it is based upon the ratio of $${C^{14}}$$ and $${C^{12}}.$$
2.
In the following radioactive decay, $$_{92}{X^{232}}{ \to _{89}}{Y^{220}},$$ how many $$\alpha $$ and $$\beta $$ -particles are ejected from $$X$$ to $$Y?$$
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.
$$_{92}{X^{232}}\frac{{ - 3\alpha \,and}}{{ - 3\beta }}{\,_{89}}{Y^{220}}$$
3.
Number of neutrons in a parent nucleus $$X,$$ which gives $$_7{N^{14}}$$ nucleus after two successive $$\beta $$ - emissions would be
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
4.
A nuclide of an alkaline earth metal undergoes radioactive decay by emission of three $$\alpha $$ - particles in succession. The group of the periodic table to which the resulting daughter element would belong is
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.
5.
In a radioactive decay, an emitted electron comes from
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
$$_0{n^1} \to {\,_1}{H^1} + {\,_{ - 1}}{e^0}$$
6.
If $$_b{X^a}$$ species emit firstly a positron, then two $$\alpha $$ and two $$\beta $$ and in last one $$\alpha $$ is also emitted and finally convert in $$_d{Y^c}$$ species, so correct the relation is
$$\eqalign{
& {\text{Hence,}}\,\,d = b - 5,\,c = a - 12 \cr
& {\text{or}}\,\,\,\,\,\,\,\,\,\,\,\,\,a = c + 12,\,b = d + 5 \cr} $$
7.
Sulphur $$ = 35\left( {34.96903\,u} \right)$$ emits a $$\beta $$ - particle but no $$\gamma - ray.$$ The product is chlorine $$ = 35\,\left( {34.96885\,u} \right).$$ The maximum energy emitted by the $$\beta $$ - particle is
10.
The half-life of a radioactive isotope is $$3 h.$$ If the initial mass of the isotope was $$300$$ $$g,$$ the mass which remained undecayed after $$18$$ $$h$$ would be