Nuclei

The $\mathrm{\beta }-\mathrm{decay}$ of ${}_{10}{}^{23}\mathrm{Ne}$ may be represented as:

            ${}_{10}{}^{23}\mathrm{Ne} \to {}_{11}{}^{23}\mathrm{Na} - {}_{-1}{}^0\mathrm{e} + \overline{\mathrm{v}} + \mathrm{Q}$ 

Ignoring the rest mass of antineutrino $\left(\overline{\mathrm{v}}\right)$ and electron , we get 

Mass defect,                  
                   $$\begin{gathered} ∆\mathrm{m} = \mathrm{m}\left({}_{10}{}^{23}\mathrm{Ne}\right) - \mathrm{m} \left({}_{11}{}^{23}\mathrm{Na}\right) \\ = 22.994466 - 22.989770 \\ = 0.004696 \mathrm{u} \end{gathered}$$ 

$\therefore$  $\mathrm{Q} = 0.004696\times 931 \mathrm{MeV} = 4.372 \mathrm{MeV}.$ 

This energy of 4.3792 MeV, is shared by ${\mathrm{e}}^{-} \mathrm{and} \overline{\mathrm{v}}$ pair because, ${}_{11}{}^{23}\mathrm{Na}$ is very massive.

The maximum K.E. of ${\mathrm{e}}^{-} = 4.372 \mathrm{MeV},$ when energy carried by $\overline{\mathrm{v}}$ is zero.

Given,
Half- life of ${}_{38}{}^{90}\mathrm{Sr}$ = 28 years. 

Using the formula,
                           $\boxed{\mathrm{\lambda } =\frac{0.693}{\mathrm{T}}}$

$\Rightarrow$                       $\mathrm{\lambda } = \frac{0.693}{28\times 365\times 24\times 60\times 60}$
                                  
                              $= 7.85 \times {10}^{-10} {\mathrm{s}}^{-1}$ 
90 g of Sr contains 6.023 x 10²³ atoms.

$\therefore$ 15 mg of Sr contains,

${\mathrm{N}}_0 = \frac{6.023 \times {10}^{23} \times 15 \times {10}^{-3}}{90}\mathrm{atoms}$ 

${\mathrm{N}}_0 = 1.0038 \times {10}^{20} \mathrm{atoms}$

Disintegration rate, $\boxed{\frac{\mathrm{dN}}{\mathrm{dt}} = -\mathrm{\lambda } {\mathrm{N}}_0}$

                                       $$\begin{gathered} = -7.85 \times {10}^{-10} \times 1.0038 \times {10}^{20} \\ = 7.88 \times {10}^{10} \mathrm{dps} \mathrm{or} \mathrm{Bq} \\ = \frac{7.88 \times {10}^{10}}{3.7 \times {10}^{10}}\mathrm{Ci} \end{gathered}$$

                                    $= 2.13 \mathrm{Ci}.$

For the given reaction, mass defect is, 
                               $$\begin{gathered} ∆m = [\mathrm{m} \left({}_6{}^{11}\mathrm{C} - 6{\mathrm{m}}_{\mathrm{e}}\right) - \left[\mathrm{m} \left({}_5{}^{11}\mathrm{B}\right) - 5 {\mathrm{m}}_{\mathrm{e}}+{\mathrm{m}}_{\mathrm{e}}\right] \\ = \mathrm{m}\left({}_6{}^{11}\mathrm{C}\right) - \mathrm{m}({}_5{}^{11}\mathrm{B}) - 2 {\mathrm{m}}_{\mathrm{e}} \\ = 11.011434 \mathrm{u} - 11.009305 \mathrm{u} - 2 \times 0.000548 \mathrm{u} \\ = 0.001033 \mathrm{u} \end{gathered}$$

Now, Q-value is , 

              $$\begin{gathered} \mathrm{Q} = 0.001033 \times 931.5 \mathrm{MeV} \\ = 0.962 \mathrm{MeV} \end{gathered}$$ 

which, is the maximum energy of the positron. 
We have, 
                     $\mathrm{Q} = {\mathrm{E}}_{\mathrm{d}}+{\mathrm{E}}_{\mathrm{e}}+{\mathrm{E}}_{\mathrm{v}}$ 
The daughter nucleus is too heavy compared to e⁺ and v. So, it carries negligible energy (E_d ≈ 0). If the kinetic energy (E_v) carried by the neutrino is minimum (i.e., zero), the positron carries maximum energy, and this is practically all energy Q.
Hence, maximum E_e≈ Q.

Given, mass of the coin = 3.0 g 

Mass of atom = 62.92960 u

Each atom of the copper contains 29 protons and 34 neutrons.

Mass of 29 electrons = 29 x 0.000548 u
                                = 0.015892 u

Mass of nucleus  = (62.92960 - 0.015892) u

                          = 62.913708 u

Mass of 29 protons = 29 x 1.007825 u
                             = 29.226925 u 

Mass of 34 neutrons = 34 x 1.008665 u 

                                = 34.29461 u
Total mass of protons and neutrons  = (29.226925 + 34.29461) u
                                                       = 63.521535 u

Binding energy  = (63.521535 - 62.913708) x 931.5 MeV
                        = 0.607827 x 931.5 MeV 

$\therefore$ Required energy = $\frac{6.023 \times {10}^{23}}{63}\times 3\times 0.607827\times 931.5 \mathrm{MeV}$
                              $= 1.6 \times {10}^{25} \mathrm{MeV} = 2.6 \times {10}^{12}\mathrm{J}.$

Using the relation between the radius of nucleus and atomic mass,
                             $\boxed{\mathrm{R} \approx {\mathrm{A}}^{1/3}}$ 

Atomic mass of gold, A₁ = 197 
Atomic mass of silver, A₂ = 107

$\therefore$                       $\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2} = {\left(\frac{{\mathrm{A}}_1}{{\mathrm{A}}_2}\right)}^{1/3}$ 

                                $= {\left(\frac{197}{107}\right)}^{1/3} = (1.84{)}^{1/3}$ 

Now, taking log on both sides, 

$\Rightarrow$         ${\log }_{10}\left(\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}\right) = {\log }_{10}(1.84{)}^{1/3}$ 

$\Rightarrow$         ${\log }_{10}\left(\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}\right) = \frac{1}{3}{\log }_{10}(1.84)$ 

                              $$\begin{gathered} = \frac{1}{3}\times 0.2648 \\ = 0.08827 \end{gathered}$$ 

$\Rightarrow$                   $\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2} = \mathrm{antilog}(0.08827)$
                              = 1.23, which is the required ratio of the nucleii.