91影视

The given data is listed below as,

• The proton鈥檚 mass is,$m_p=1.0073\mathrm{u}$
• The deuteron鈥檚 mass is,$m_d=20136\mathrm{u}$
• The helium-nucleus鈥檚 mass is, $m_{He}=3.0155\mathrm{u}$

• The proton or neutron鈥檚 approximate radius is,$r_p=r_d=1脳{10}^{-15}\mathbf{m}$

The law of the energy conservation states that an isolated system鈥檚 total energy remains constant and it is conserved with time

The concept of the law of energy conservation gives the kinetic energies of the proton and the helium-3 nucleus along with the net energy release and the kinetic energy generated.

The equation of the initial kinetic energy of the proton and the deuteron can be expressed as:

$m_p{c}^2+m_d{c}^{2+}{k}_p+k_d=\frac{1}{4蟺{蔚}_0}-\frac{e^2}{r}+m_d{c}^2$

Here, $m_p$is the mass of the proton, C is the speed of light, $m_d$is the mass of the deuteron ,$k_p$is the kinetic energy of the proton ,$k_d$is the kinetic energy of the deuteron,$\frac{1}{4蟺{蔚}_0}$is the Coulomb鈥檚 constant,e is the charge on the electron and r is the combined radius of the proton and the deuteron.

The above equation can be reduced as follows,

$k_p+k_d=\frac{1}{4蟺{蔚}_0}-\frac{e^2}{r}$

Substitute $9脳{10}^9{\mathrm{Nm}}^2/{\mathrm{C}}^2\mathrm{for}\frac{1}{4{\mathrm{蟺蔚}}_0},1.602脳{10}^{-19}\mathrm{C}\mathrm{for}\mathrm{e},\mathrm{and}2脳{10}^{-15}\mathrm{m}\mathrm{for}r$in the above expression.

$$\begin{gathered} {\mathrm{k}}_{\mathrm{p}}+{\mathrm{k}}_{\mathrm{d}}=\left(9脳{10}^9{\mathrm{Nm}}^2/{\mathrm{C}}^2\right)脳\frac{{\left(1.602脳{10}^{-19}\mathrm{C}\right)}^2}{2脳{10}^{-15}\mathrm{m}} \\ =1.154脳{10}^{-13}\mathrm{N}-\mathrm{m} \\ =1.154脳{10}^{-13}\mathrm{N}-\mathrm{m}脳\frac{1\mathrm{J}}{1\mathrm{N}-\mathrm{m}} \\ =1.154脳{10}^{-13}\mathrm{J} \end{gathered}$$

Covert the total initial kinetic energy of the proton and the deuteron from joules to electron volts.

$$\begin{gathered} k_p+k_d=1.154脳{10}^{-13}\mathrm{J}脳\frac{1\mathrm{eV}}{1.6脳{10}^{-19}\mathrm{J}} \\ =7.2脳{10}^5\mathrm{eV} \end{gathered}$$

Thus, the initial total kinetic energy of the proton and the deuteron in joules is$1.154脳{10}^{-13}\mathrm{J}$ and electron volts is .$7.2脳{10}^5\mathrm{eV}$

The equation of the total energy after fusion is expressed as,

$T=k_{He}+m_{He}{c}^2+k_y$

Here,$k_{He}$is the kinetic energy of the helium,$k_y$is the kinetic energy of the gamma rays and $m_{He}$is the mass of the helium-3 nucleus.

The above equation can be written as,

$m_p{c}^2+m_p{c}^2+k_p+k_d=k_{He}+m_p{c}^2+k_y$

Substitute all the values in above equation.

$$\begin{gathered} \left(10073\mathrm{u}\right)脳c^2+\left(2.0136\mathrm{u}脳\right)c^2+1.154脳{10}^{-13}\mathrm{J}={\mathrm{k}}_{\mathrm{He}}+\left(3.0155\mathrm{u}\right){\mathrm{c}}^2+{\mathrm{k}}_{\mathrm{y}} \\ {\mathrm{k}}_{\mathrm{He}}+{\mathrm{k}}_{\mathrm{y}}=\left(0.0054\mathrm{u}\right)脳{\mathrm{c}}^2+1.154脳{10}^{-13}\mathrm{J} \\ =\left[\begin{array}{c}\left(0.0054\mathrm{u}脳\left(\frac{1.166脳{10}^{-27}\mathrm{kg}}{1\mathrm{u}}\right)\right)脳3脳{10}^8\mathrm{m}/\mathrm{s}\\ +1.154脳{10}^{-13}\mathrm{J}\\ \end{array}\right] \\ =8.06脳{10}^{-13}\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2+1.154脳{10}^{-13}\mathrm{J} \\ =\left[\begin{array}{c}8.06脳{10}^{-13}\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2脳\frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\\ +1.154脳{10}^{-13}\mathrm{J}\\ \end{array}\right] \\ =8.06脳{10}^{-13}\mathrm{J}+1.154脳{10}^{-13}\mathrm{J} \\ =9.23脳{10}^{-13}\mathrm{J} \end{gathered}$$Convert the total kinetic energy of the helium and the gamma ray from joules to electron volts.

$$\begin{gathered} k_{He}+k_y=9.23脳{10}^{-13}J脳\frac{1eV}{1.602脳{10}^{-19}J} \\ =5.76脳{10}^{6}eV \end{gathered}$$

Thus, the kinetic energy of the helium-3 nucleus and the energy of the gamma ray in joules is $9.23脳{10}^{-13}\mathrm{J}$and electro volt is $5.76脳{10}^6\mathrm{eV}$.

The expression of the net energy release is expressed as follows,

$Q=\left(k_{He}+k_y\right)-\left(k_p+k_d\right)$

Substitute all the values in above expression.

$$\begin{gathered} Q=9.23脳{10}^{-13}\mathrm{J}-1.154脳{10}^{-13}\mathrm{J} \\ =8.076脳{10}^{-13}\mathrm{J} \end{gathered}$$

The net energy release in electro volt is expressed as-

$$\begin{gathered} Q=8.076脳{10}^{-13}\mathrm{J}脳\frac{1\mathrm{eV}}{1.602脳{10}^{-19}\mathrm{J}} \\ =5.04脳{10}^6\mathrm{eV} \end{gathered}$$

Thus, the net energy release in joules and electron volts are $8.076脳{10}^{-13}\mathrm{J}$and $5.04脳{10}^6\mathrm{eV}$respectively.

The expression for the kinetic energy generated is expressed as follows,

$K.E=\left(k_{He}+k_y\right)脳N_A$

Here,$N_A=6.023脳{10}^{23}$is the Avogadro鈥檚 number

Substitute all the values in the above expression.

$K.E.=\left(9.23脳{10}^{-13}\mathrm{J}\right)脳6.023脳{10}^{23}$

Thus, the kinetic energy generated is $5.56脳{10}^{11}\mathrm{J}$.

As the potential energy curve shows the potential energy of the fusion reaction in different position, hence curve 2 shows that the potential energy decreases with the increase in the radius.

Thus, the 2^nd potential energy curve is a reasonable representation of the interaction in this fusion reaction.