Nuclei

Initial total binding energy is

$\begin{array}{rl}2\times 3\times 5.6+4\times 7.4& =33.6+29.6\\ & =63.2\mathrm{MeV}\end{array}$

Final binding energy is

$10\times 6.1=61.0\mathrm{MeV}$

Therefore,

$\mathrm{\Delta }M{c}^2=63.2-61.0=2.2\mathrm{MeV}$

$n\to p+\overline{e}+\overline{v}$

For $\alpha$-decay, kinetic energy of the $\alpha$-particle is

$K_{\alpha }=\left(\frac{A-4}{A}\right)Q$

For substance A with mass number 200:

$K_{\alpha 1}=\frac{196}{200}Q$

For substance B with mass number 212:

$K_{\alpha 2}=\frac{208}{212}Q$

Therefore,

$\frac{K_{\alpha 1}}{K_{\alpha 2}}=\frac{196}{200}·\frac{212}{208}=\frac{2597}{2600}$

Binding energy per nucleon of deuteron \( \left(^2_1\text{H}\right) = 1.1\, \text{MeV} \)
Binding energy per nucleon of helium-4 \( \left(^4_2\text{He}\right) = 7.0\, \text{MeV} \)

Two deuterons \( \Rightarrow \) total nucleons = 2 × 2 = 4

\[
\text{BE}_{\text{reactants}} = 2 \times (1.1 \times 2) = 4.4\, \text{MeV}
\]

\[
\text{BE}_{\text{product}} = 4 \times 7.0 = 28.0\, \text{MeV}
\]

$$\begin{gathered} Q={\text{BE}}_{\text{product }}-{\text{BE}}_{\text{reactants }}=28.0-4.4 \\ =23.6\text{MeV} \end{gathered}$$

The number of remaining nuclei after time t can be given by:

$N\left(t\right)=N_0{e}^{-\lambda t}$

For half-life $t_{1/2}$ of $n_1$, we know that:

$t_{1/2}=\frac{\ln 2}{{\lambda }_1}$

At the end of one half-life, the number of $n_1$ nuclei left will be:

$N_1=\frac{N_0}{2}$

For nucleus $n_2$, which has a decay constant 3 times that of $n_1$ the remaining number of nuclei will be:

$N_2=N_0{e}^{-3}{\lambda }_{1}t$

Since $t=t_{1/2}=\frac{\ln 2}{{\lambda }_1}$, we substitute it into the equation for $N_2$ :

$N_2=N_0{e}^{-3{\lambda }_1\times \frac{\ln 2}{{\lambda }_1}}=N_0{e}^{-3\ln 2}=N_0{\left(\frac{1}{2}\right)}^3=\frac{N_0}{8}$

Now, the ratio of the number of nuclei of $n_2$ to $n_1$ after one half-life of $n_1$ is:

$\frac{N_2}{N_1}=\frac{\frac{N_0}{8}}{\frac{N_0}{2}}=\frac{1}{4}$

Two deuterium atoms (${}_1{}^2\text{H}$) fuse to form one helium atom (${}_2{}^4\text{He}$):

Calculate Total Binding Energies

Energy released ($\Delta E$) is the difference between the total binding energy of the products and the reactants.

Total BE of Reactants (\(2\) deuterium atoms):

Each ${}_1{}^2\text{H}$ has \(2\) nucleons.

Total BE of Product (\(1\) helium atom):

Each ${}_2{}^4\text{He}$ has \(4\) nucleons.

Calculate Energy Released ($\Delta E$)

Correct Option: (b) 24.4

Nuclear force is short range.

It is the equation of neutron (n) decays into a proton (p), in which an electron (e⁻) is emitted, and an antineutrino is emitted to conserve lepton number.

Binding energy per nucleon of deuteron \( \left(^2_1\text{H}\right) = 1.1\, \text{MeV} \)
Binding energy per nucleon of helium-4 \( \left(^4_2\text{He}\right) = 7.0\, \text{MeV} \)

Two deuterons \( \Rightarrow \) total nucleons = 2 × 2 = 4

\[
\text{BE}_{\text{reactants}} = 2 \times (1.1 \times 2) = 4.4\, \text{MeV}
\]

\[
\text{BE}_{\text{product}} = 4 \times 7.0 = 28.0\, \text{MeV}
\]

\[
Q = \text{BE}_{\text{product}} - \text{BE}_{\text{reactants}} = 28.0 - 4.4 = \boxed{23.6\, \text{MeV}}
\]

Nuclear reaction considered
${C_6}^{13}+Energy\to {C_6}^{12}+{n_0}^1$
Mass defect calculation
$\Delta M=(12.000000+1.008665)-13.003354$
Value of mass defect
$\Delta M=-0.00531$
Energy required for reaction
$E=0.00531\times 931.5$
Final energy required
$E=4.95$ MeV

Nuclear reaction considered
${C_6}^{13}+Energy\to {C_6}^{12}+{n_0}^1$
Mass defect calculation
$\Delta M=(12.000000+1.008665)-13.003354$
Value of mass defect
$\Delta M=-0.00531$
Energy required for reaction
$E=0.00531\times 931.5$
Final energy required
$E=4.95$ MeV

Mass defect:
$\Delta m=\left(6m_p+6m_n\right)-m_C$
$\Delta m=6(1.00727+1.00866)-12$
$\Delta m=6(2.01593)-12=12.09558-12$
=0.09558 u
$\Delta m=0.09558\times 931.5=89.03\mathrm{MeV}/{\mathrm{c}}^2$

Convert mass to kilograms:\(m=1\,\text{g}=1\times 10^{-3}\,\text{kg}\)

Square the speed of light:\(c^{2}=(3\times 10^{8}\,\text{m/s})^{2}=9\times 10^{16}\,\text{m}^{2}/\text{s}^{2}\)

Calculate the energy:\(E=m\times c^{2}=(1\times 10^{-3}\,\text{kg})\times (9\times 10^{16}\,\text{m}^{2}/\text{s}^{2})=9\times 10^{13}\,\text{J}\) Other Units The energy can also be expressed in other units:

Approximately \(2.5\times 10^{7}\) kilowatt-hours (kWh).Approximately \(5.6\times 10^{26}\) mega-electron volts (MeV).

Convert mass to kilograms:\(m=1\,\text{g}=1\times 10^{-3}\,\text{kg}\)

Square the speed of light:\(c^{2}=(3\times 10^{8}\,\text{m/s})^{2}=9\times 10^{16}\,\text{m}^{2}/\text{s}^{2}\)

Calculate the energy:\(E=m\times c^{2}=(1\times 10^{-3}\,\text{kg})\times (9\times 10^{16}\,\text{m}^{2}/\text{s}^{2})=9\times 10^{13}\,\text{J}\) Other Units The energy can also be expressed in other units:

Approximately \(2.5\times 10^{7}\) kilowatt-hours (kWh).Approximately \(5.6\times 10^{26}\) mega-electron volts (MeV).

$$
\begin{aligned}
&\rho=\frac{M}{V}=\frac{M_n \times A}{\frac{4}{3} \pi R^3}=\frac{m_n \times A}{\frac{4}{3} \pi A R_0^3}\\
&\text { so } \boldsymbol{\rho}_{\text {is almost is constant }}\\
&\begin{aligned}
& R=R_0 A^{1 / 3} \\
& R \propto A^{1 / 3}
\end{aligned}
\end{aligned}
$$

The number of remaining nuclei after time t can be given by:

$N\left(t\right)=N_0{e}^{-\lambda t}$

For half-life $t_{1/2}$ of $n_1$, we know that:

$t_{1/2}=\frac{\ln 2}{{\lambda }_1}$

At the end of one half-life, the number of $n_1$ nuclei left will be:

$N_1=\frac{N_0}{2}$

For nucleus $n_2$, which has a decay constant 3 times that of $n_1$ the remaining number of nuclei will be:

$N_2=N_0{e}^{-3}{\lambda }_{1}t$

Since $t=t_{1/2}=\frac{\ln 2}{{\lambda }_1}$, we substitute it into the equation for $N_2$ :

$N_2=N_0{e}^{-3{\lambda }_1\times \frac{\ln 2}{{\lambda }_1}}=N_0{e}^{-3\ln 2}=N_0{\left(\frac{1}{2}\right)}^3=\frac{N_0}{8}$

Now, the ratio of the number of nuclei of $n_2$ to $n_1$ after one half-life of $n_1$ is:

$\frac{N_2}{N_1}=\frac{\frac{N_0}{8}}{\frac{N_0}{2}}=\frac{1}{4}$

$n\to p+\overline{e}+\overline{v}$

It is the equation of neutron (n) decays into a proton (p), in which an electron (e⁻) is emitted, and an antineutrino is emitted to conserve lepton number.

$\begin{array}{l}R=R_0{A}^{\frac{1}{3}}\\ \mathrm{Density}=\frac{\mathrm{Mass} \mathrm{number}}{\mathrm{Volume}}\\ =\frac{A}{\frac{4}{3}\pi {R_0}_3^{ }A}\\ =\mathrm{constant}\end{array}$

Mass defect: $\Delta m=\left[Z{m}_p+(A-Z)m_n\right]-m_{nucleus}$

$= \left[83\right(1.007825) + 126(1.008665\left)\right] – 208.980388$

$= (83.649 + 127.092) – 208.980388 \approx 1.7606u$

Binding energy $(B=\Delta m\times 931)$:

$B = 1.7606 \times 931 \approx 1639 MeV$

Binding energy per nucleon: $\frac{B}{A}=\frac{1639}{209}\approx 7.84MeV$