91影视

The mass is given by,

$$\begin{gathered} \mathit{m}\mathbf{=}\mathit{蚁}\mathit{V} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}}{\mathbf{蟺谤}}^{\mathbf{3}}\mathbf{蚁} \end{gathered}$$

Here, p is density, and V is volume.

There are equal numbers of ${}^{\mathbf{2}}\mathbf{H}$and${}^{\mathbf{3}}\mathbf{H}$present, therefore the expression for the number of ${}^{\mathbf{2}}\mathbf{H}$and${}^{\mathbf{3}}\mathbf{H}$is given by,

${\mathbf{N}}_{\mathbf{2}\mathbf{H}}\mathbf{=}{\mathbf{N}}_{\mathbf{2}\mathbf{H}}\mathbf{=}\frac{{\mathbf{N}}_{\mathbf{A}}\mathbf{m}}{{\mathbf{M}}_{\mathbf{2}\mathbf{H}}\mathbf{+}{\mathbf{M}}_{\mathbf{3}\mathbf{H}}}$

Each fusion reaction releases Q=17.59 MeV of energy, with 10% efficiency, then the total energy released by the pellet is given by,

$$\begin{gathered} \mathit{E}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}{\mathit{N}}_{\mathbf{2}\mathbf{H}}\mathit{Q} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{10}\left(\frac{N_{A}m}{M_{2H}+M_{3H}}\right)\mathit{Q} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{4}\left(0.10\right){\mathbf{蟺谤}}^{\mathbf{3}}\mathbf{}{\mathbf{蚁N}}_{\mathbf{A}}\mathbf{Q}}{\mathbf{3}\left(M_{2H}+M_{3H}\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\left(1\right) \end{gathered}$$

(a)

The molar mass $M_{2H}$of deuterium atoms is $2.0脳{10}^{-3}\mathrm{kg}/\mathrm{mol}$, and the molar mass $m_{3H}$of tritium atoms is $3.0脳{10}^{-3}\mathrm{kg}/\mathrm{mol}$. The uncompressed radius of the fuel pellet is $20渭m$and its density is $蚁=200\mathrm{kg}/{\mathrm{m}}^3$.

Substitute all the known values in equation (1).

$$\begin{gathered} E=\frac{4\left(0.10\right)\mathrm{蟺}{\left(20脳{10}^{-6}\mathrm{m}\right)}^3\left(200\mathrm{kg}/{\mathrm{m}}^3\right)\left(6.022脳{10}^{23}{\mathrm{mol}}^{-1}\right)\left(17.59\mathrm{MeV}\right)}{3\left(2.0脳{10}^{-3}\mathrm{kg}/\mathrm{mol}+3.0脳{10}^{-3}\mathrm{kg}/\mathrm{mol}\right)} \\ =1.42脳{10}^{15}\mathrm{MeV} \\ =\left(1.421脳{10}^{15}\mathrm{Mev}\right)\left(1.602脳{10}^{13}\mathrm{J}/\mathrm{MeV}\right) \\ 鈮�227\mathrm{J} \end{gathered}$$

Therefore, the energy released in micro-explosionis 227 J.

(b)

The expression to calculate the amount of TNT is given by,

$m=\frac{E}{E_{TNT}}......\left(2\right)$

Substitute all the known values in equation (2).

$$\begin{gathered} m=\frac{227\mathrm{J}}{4.6脳{10}^6\mathrm{J}} \\ =4.93脳{10}^{-5}\mathrm{kg} \\ =49.3\mathrm{mg} \end{gathered}$$

Therefore, the amount of TNT required is 49.3 mg.

(c)

The expression to calculate power generated is given by,

$P=\left(\frac{dN}{dt}\right)E.......\left(3\right)$

Substitute all the known values in equation (3).

$$\begin{gathered} P=\left(100s^{-1}\right)\left(227J\right) \\ =22700\mathrm{W} \\ =22.7\mathrm{KW} \end{gathered}$$

Therefore, the power generated is 22.7 kW.