51.
The electrostatic energy of $$Z$$ protons uniformly distributed throughout a spherical nucleus of radius $$R$$ is given by
$$E = \frac{3}{5}\frac{{Z\left( {Z - 1} \right){e^2}}}{{4\pi {\varepsilon _0}R}}$$
The measured masses of the neutron $$_1^1H,\,_7^{15}N$$ and $$_8^{15}O$$ are $$1.008665\,u,\,1.007825\,u,\,15.000109\,u$$ and $$15.003065\,u,$$ respectively. Given that the radii of both the $$_7^{15}N$$ and $$_8^{15}O$$ nuclei are same, $$1\,u = 931.5\,Me\,V/{c^2}$$ ($$e$$ is the speed of light) and $$\frac{{{e^2}}}{{\left( {4\pi {\varepsilon _0}} \right)}} = 1.44\,MeV\,fm.$$ Assuming that the difference between the binding energies of $$_7^{15}N$$ and $$_8^{15}O$$ is purely due to the electrostatic energy, the radius of either of the nuclei is
$$\left( {1fm. = {{10}^{ - 15}}m} \right)$$
52.
A radioactive clement decays by $$\beta $$ emission. A detector records $$n$$ beta particles in 2 second and in next 2 seconds it records $$0.75\,n$$ beta particles. Find mean life corrected to nearest whole number. Given $$\ell n2 = 0.6931$$ and $$\ell n3 = 1.0986.$$
We know that, $$N = {N_0}{e^{ - \lambda t}}$$
After 2 second, $${N_1} = {N_0}{e^{ - \lambda \times 2}}$$
After $$\left( {2 + 2} \right)$$ second, $${N_2} = {N_0}{e^{ - \lambda \times 4}}$$
According to given conditions,
$$\eqalign{
& {N_0} - {N_1} = n \cr
& {\text{or}}\,\,{N_0} - {N_0}{e^{ - 2\lambda }} = n\,......\left( {\text{i}} \right) \cr
& {\text{and}}\,{N_0}{e^{ - 2\lambda }} - {N_0}{e^{ - 4\lambda }} = 0.75\,n\,......\left( {{\text{ii}}} \right) \cr} $$
After solving above equations, we get
$$\lambda = 0.145\,s$$
Mean life$$T = \frac{1}{\lambda } = 6.9\,s.$$
53.
In a radioactive sample, $$_{19}^{40}K$$ nuclei either decay into stable $$_{20}^{40}Ca$$ nuclei with decay constant $$4.5 \times {10^{10}}$$ per year or into stable $$_{18}^{40}Ar$$ nuclei with decay constant $$0.5 \times {10^{ - 10}}$$ per year. Given that in this sample all the stable $$_{20}^{40}Ca$$ and $$_{18}^{40}Ar$$ nuclei are produced by the $$_{19}^{40}K$$ nuclei only. In time $$t \times {10^9}$$ years, if the ratio of the sum of stable $$_{20}^{40}Ca$$ and $$_{20}^{40}Ca$$ nuclei to the radioactive $$_{19}^{40}K$$ nuclei is 99, the value of $$t$$ will be [Given: $$l$$n $$10 = 2.3$$ ]
54.
The half-life of a radioactive substance is 30 minutes. The time (in minutes) taken between $$40\% $$ decay and $$85\% $$ decay of the same radioactive substance is
Key Idea
Half-life of a radioactive substance is $${T_{\frac{1}{2}}} \propto \log \left( {\frac{{{N_0}}}{N}} \right)$$
Given, $${N_1} = 0.6\,{N_0}\,\,\left( {\because 40\% \,{\text{decay}}} \right)$$
$${N_2} = 0.15\,{N_0}\,\,\left( {\because 85\% \,{\text{decay}}} \right)$$
Putting these in the formula,
$$\frac{{{N_2}}}{{{N_1}}} = \frac{{0.15\,{N_0}}}{{0.6\,{N_0}}} = \frac{1}{4} = {\left( {\frac{1}{4}} \right)^2}$$
So, two half-life periods has passed.
Thus, time taken $$ = 2 \times {t_{\frac{1}{2}}} = 2 \times 30 = 60\,\min $$
55.
The half-life of a radioactive isotope $$X$$ is $$20\,yr.$$ It decays to another element $$Y$$ which is stable. The two elements $$X$$ and $$Y$$ were found to be in the ratio $$1:7$$ in a sample of a given rock. The age of the rock is estimated to be
As we know that
$${N = {N_0}{{\left( {\frac{1}{2}} \right)}^n}}$$
$$\frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^3} = \frac{1}{{1 + 7}} = \frac{1}{8}$$
So, number of half lifes = 3
$$\eqalign{
& \Rightarrow T = 20\,yr \cr
& \therefore T = \frac{t}{n} \cr
& \Rightarrow t = Tn = 20 \times 3\,yr = 60\,yr \cr} $$
56.
If $${M_O}$$ is the mass of an oxygen isotope $$_8{O^{17}},{M_P}$$ and $${M_N}$$ are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is
A
$$\left( {{M_O} - 17{M_N}} \right){c^2}$$
B
$$\left( {{M_O} - 8{M_P}} \right){c^2}$$
C
$$\left( {{M_O} - 8{M_P} - 9{M_N}} \right){c^2}$$
Number of atoms left after $$n$$ half-lives is given by
$$N = {N_0}{\left( {\frac{1}{2}} \right)^n}\,\,{\text{or}}\,\,\frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^n}$$
Number of half-lives,
$$\eqalign{
& n = \frac{t}{T} = \frac{{6400}}{{800}} = 8 \cr
& \therefore \frac{N}{{{N_0}}} = {\left( {\frac{1}{2}} \right)^8} = \frac{1}{{256}} \cr} $$ Alternative
Let the initial part be unity
So, after 800 years, it will remain $$ = \frac{1}{2}$$
after 1600 years, it will remain $$ = \frac{1}{4}$$
after 2400 years, it will remain $$ = \frac{1}{8}$$
after 3200 years, it will remain $$ = \frac{1}{16}$$
after 4000 years, it will remain $$ = \frac{1}{32}$$
after 4800 years, it will remain $$ = \frac{1}{64}$$
after 5600 years, it will remain $$ = \frac{1}{128}$$
after 6400 years, it will remain $$ = \frac{1}{256}$$
58.
Half lives of two isotopes $$X$$ and $$Y$$ of a material are known to be $$2 \times {10^9}{\text{years}}$$ and $$4 \times {10^9}{\text{years}}$$ respectively. If a planet was formed with equal number of these isotopes, then the current age of planet, given that currently the material has $$20\% $$ of $$X$$ and $$80\% $$ of $$Y$$ by number, will be -
59.
A sample of radioactive material $$A,$$ that has an activity of $$10\,mCi\left( {1\,Ci3.7 \times {{10}^{10}}\,decays/s} \right),$$ has twice the number of nuclei as another sample of a different radioactive material $$B$$ which has an acitvity of $$20\,mCi.$$ The correct choices for half-lives of $$A$$ and $$B$$ would then be respectively: