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In any radioactive sample, undergoing $\mathrm{\alpha }, \mathrm{\beta } \mathrm{or} \mathrm{\gamma }$decay, the number of nucleii undergoing decay per unit time is directly proportional to the total number of nuclei in the sample. This is known as the radioactive decay law. 

Let, N be the number of nucleii in the sample, 
$∆$N is the sample undergoing decay and, 
$∆\mathrm{t}$ is the time then, 

                            $\frac{∆\mathrm{N}}{∆t} \propto N$ 

$\Rightarrow$                        $\frac{∆\mathrm{N}}{∆\mathrm{t}}= \lambda N$ 

where, $\mathrm{\lambda }$ is the decay constant. 


Numerical: 

Half -period of radioactive substance = 30 days
Number of atoms disintegrated = $\left(\frac{3}{4}\right) {\mathrm{N}}_0$
Number of atoms left after time t,  $\mathrm{N} = {\mathrm{N}}_0 - \frac{3}{4}{\mathrm{N}}_0 = \frac{1}{4}{\mathrm{N}}_0$ 

Number of half lives in time t days,
                                    
                        $\boxed{\mathrm{n} = \frac{\mathrm{t}}{\mathrm{T}}} = \frac{\mathrm{t}}{30}$
where, 

T = Half life time
n = no. of half lives
t = time for disintegrates 

Number of nuclei left after n half lives is given by, 

                      $\boxed{\mathrm{N} = {\mathrm{N}}_0{\left(\frac{1}{2}\right)}^{\mathrm{N}}}$ 
Therefore, 
                    $\frac{{\mathrm{N}}_0}{4} = {\mathrm{N}}_0{\left(\frac{1}{2}\right)}^{\mathrm{t}/30}$

$\Rightarrow$           $(2{)}^{\mathrm{t}/30} = 4 = (2{)}^2$

$\Rightarrow$               $\frac{\mathrm{t}}{30} = 2 \mathrm{or} \mathrm{t} = 60 \mathrm{days}$

(ii) Now, using the formula,  $\boxed{\mathrm{N} = {\mathrm{N}}_0{\left(\frac{1}{2}\right)}^{\mathrm{n}}}$

$\therefore$                    $\frac{{\mathrm{N}}_0}{8} = {\mathrm{N}}_0{\left(\frac{1}{2}\right)}^{\mathrm{t}/30}$

$\Rightarrow$               $(2{)}^{\mathrm{t}/30} = 8 = (2{)}^3$      
      
$\Rightarrow$                  $\frac{\mathrm{t}}{30} = 3$
i.e.,                      $\mathrm{t} = 90 \mathrm{days}$, is the time taken for 1/8 of the original number of atoms to remain unchanged.

                                        
                                    

 

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Classification of nuclides is as follows: 

Isotopes: ${}_{ 6}{}^{12}\mathrm{C}, {}_6{}^{14}\mathrm{C}$ ; has the same atomic number but, different mass number. 

Isobars $: {}_2{}^3\mathrm{He}, {}_1{}^3\mathrm{H}$ ; has the same mass number but, different atomic number.

Isotones: ${}_{80}{}^{198} \mathrm{Hg}, {}_{79}{}^{197} \mathrm{Au}$ ; has same neutron number but, different proton number. 

Now, 

Radius of nucleus, $\boxed{\mathrm{R} = {\mathrm{R}}_0 {\mathrm{A}}^{1/3}}$                          ...(i) 
where,  ${\mathrm{R}}_0$is the range of nuclear force (or Nuclear Unit Radius) 

Density is given by, $\mathrm{\rho } = \frac{\mathrm{M}}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3}$                        ...(ii) 

But,                         $\mathrm{M}\propto \mathrm{A}$                               ...(iii) 

From (i) and (iii), we get

$\mathrm{R}\propto {\mathrm{M}}^{1/3} \Rightarrow {\mathrm{R}}^3 \propto \mathrm{M} \Rightarrow {\mathrm{R}}^3 = \mathrm{kM}$ where,

$\mathrm{k}\to \mathrm{proportionality} \mathrm{constant}$                                 ...(iv) 

From (ii) and (iv), we have
                                   $\mathrm{\rho } = \frac{\mathrm{M}}{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3}= \frac{3}{4\mathrm{\pi }}\frac{\mathrm{M}}{\mathrm{KM}} = \mathrm{constant}.$

Thus, we can see that nuclear density is clearly independent of mass number A. 

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Decay constant of a radioactive element is the reciprocal of the time during which the number of atoms left in the sample reduces to $\frac{1}{\mathrm{e}}$ times the number of atoms in the original sample. 

Given, two nucleii  P and Q. 
P and Q have equal number of atoms at t=0. 
Half life of P = 3 hours. 
Half life of Q = 9 hours. 

and, t = 18 hours

Number of half lives of P in 18 hours  =$\frac{\mathrm{t}}{{\mathrm{T}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$= $\frac{18}{3}=6$ 

Number of nuclei P left undecayed after 6 half lives is, 
                      ${\mathrm{N}}_1 = \mathrm{N}{\left(\frac{1}{2}\right)}^6$ 

Number of half lives of Q in 18 hours=$\frac{\mathrm{t}}{{\mathrm{T}}_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$ = $\frac{18}{9}=2$ 

Number of nuclei of Q left undecayed after 2 half lives is, 

                       ${\mathrm{N}}_2 = \mathrm{N}{\left(\frac{1}{2}\right)}^2$ 

Ratio of their decay rate is, 

               $\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2} = \frac{{\mathrm{\lambda }}_1 {\mathrm{N}}_1}{{\mathrm{\lambda }}_2 {\mathrm{N}}_2} = \frac{{\mathrm{T}}_2 {\mathrm{N}}_1}{{\mathrm{T}}_1 {\mathrm{N}}_2}$ 

 $\left[\because \mathrm{R} = \mathrm{\lambda } \mathrm{N} \mathrm{and} \mathrm{T} = \frac{0.693}{\mathrm{\lambda }}\right]$ 

$\therefore$               $\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2} = \frac{9}{3}\times \frac{\mathrm{N}{\left({\displaystyle \frac{1}{2}}\right)}^6}{\mathrm{N}{\left({\displaystyle \frac{1}{2}}\right)}^2} = \frac{3}{16}$ 

Hence ratio of disintegration is, ${\mathrm{R}}_1:{\mathrm{R}}_2 = 3 :16$

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We have, 

Rate of increase of element = $\frac{\mathrm{number} \mathrm{of} \mathrm{nuclei} \mathrm{by} \mathrm{reactor}}{1 \mathrm{second}}-\frac{\mathrm{number} \mathrm{of} \mathrm{nuclei} \mathrm{decaying}}{1 \mathrm{second}}$

That is, 
                 $\frac{\mathrm{dN}}{\mathrm{dt}} = \mathrm{R} - \mathrm{\lambda N} \mathrm{or} \frac{\mathrm{dN}}{\mathrm{dt}}+\mathrm{\lambda N} = \mathrm{R}$ 
                      
The solution to this is the sum of the homogeneous solution, 
              ${\mathrm{N}}_{\mathrm{h}} = {\mathrm{ce}}^{-\mathrm{\lambda t}},$ where c is a constant, and

a particular solution, ${\mathrm{N}}_{\mathrm{l}} = \frac{\mathrm{R}}{\mathrm{\lambda }}.$ 

Therefore, the required solution is, 

                       $\mathrm{N} = {\mathrm{N}}_{\mathrm{h}}+{\mathrm{N}}_{\mathrm{p}} = {\mathrm{ce}}^{-\mathrm{\lambda t}}+\frac{\mathrm{R}}{\mathrm{\lambda }}$ 

The constant c is obtained from the requirement that the initial number of nuclei be zero, 
                           $\mathrm{N}\left(0\right) = 0 = \mathrm{c}+\frac{\mathrm{R}}{\mathrm{\lambda }}$
$\Rightarrow \mathrm{c} = -\frac{\mathrm{R}}{\mathrm{\lambda }}$
so that,                $\mathrm{N} = \frac{\mathrm{R}}{\mathrm{\lambda }}(1-{\mathrm{e}}^{-\mathrm{\lambda t}})$ 

The equilibrium value is $(\mathrm{t}\to \infty ) = \mathrm{R}/\mathrm{\lambda }.$ 

Setting N equal to 1/2 of this value gives, 

                  $\frac{1}{2}\left(\frac{\mathrm{R}}{\mathrm{\lambda }}\right) = \frac{\mathrm{R}}{\mathrm{\lambda }}(1-{\mathrm{e}}^{-\mathrm{\lambda t}})$ 

                       ${\mathrm{e}}^{-\mathrm{\lambda t}} = \frac{1}{2}$

$\Rightarrow$         $\mathrm{t} = \frac{\ln 2}{\mathrm{\lambda }} = {\mathrm{T}}_{1/2}$ 

The result is independent of R.

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Nuclear force is a strong attractive force acting between the nucleons of the atomic nucleus that holds the nucleus together. 

Characteristics of nuclear force : 

i) Nuclear forces are short range forces operating upto a distance which is of the order of a few fermi. 

ii) Nuclear forces are strongest force in nature. It's magnitude is 100 times that of the electrostatic force and 10³⁸ times that of the gravitational force. 

Numerical:

The volume  V of lake is, $$\begin{gathered} \mathrm{V} = 2.56 \times {10}^5 \times {10}^6 \times 80 {\mathrm{m}}^3 \\ = 2.048 \times {10}^{13} {\mathrm{m}}^3 \end{gathered}$$ 

$\therefore$ Mass of water, M in lake  = 2.048 x 10¹³ x 10³
                                          = 2.048 x 10¹⁶ kg
                                          = 2.048 x 10¹⁹ gm 

Now, 18 g of water contains 2 g of hydrogen. 

$\therefore$ Mass of hydrogen atom in lake = $\frac{1}{9}\times 2.048\times {10}^{19}\mathrm{g}$ 

$\therefore$ Number of atoms of hydrogen in lake = 
$6.023 \times {10}^{23} \times \frac{2.048 \times {10}^{19}}{9}$
                                                               $= 1.37 \times {10}^{42}$ 

Since abundance of ${}_1{}^2\mathrm{H}$ is only 0.0156% of hydrogen atoms, the number of ${}_1{}^2\mathrm{H}$ atoms in lake, 

$$\begin{gathered} = 1.37 \times {10}^{42} \times 1.56 \times {10}^{-4} \\ = 2.137 \times {10}^{38} \end{gathered}$$ 

Energy released due to fusion of one atom of   ${}_1{}^2\mathrm{H} = 7.17 \mathrm{MeV}$ 

$\therefore$ Energy released when all ${}_1{}^2\mathrm{H}$ atoms present undergo fusion, 

              $$\begin{gathered} = 2.137 \times {10}^{38} \times 7.17 \mathrm{MeV} \\ = 1.532 \times {10}^{39} \mathrm{MeV}. \end{gathered}$$