91Ó°ÊÓ

The disintegrated energy, $Q=170\mathrm{MeV}$

In the fission process of ${}^{235}U$, on fragment is ${}^{83}Ge$.

The expression to calculate the total energy is given as follows.

$\mathbf{Q}\mathbf{=}{\mathbf{KE}}_{\mathbf{Ge}}\mathbf{+}{\mathbf{KE}}_{\mathbf{X}}$

Here ${\mathbf{KE}}_{\mathbf{Ge}}$is the kinetic energy of the Ge and ${\mathbf{KE}}_{\mathbf{X}}$is the kinetic energy of the other fragment.

The expression to calculate the speed of fragment is given as follows.

$\mathbf{v}\mathbf{=}\sqrt{\frac{\mathbf{2}\mathbf{K}\mathbf{.}\mathbf{E}}{\mathbf{M}}}$

Here, K.E is the energy and M is the molar mass of the fragment.

It is given that no neutron is emitted and one fragment is ${}^{83}U$.

Consider the reaction given below.

${}^{235}U_{92}+{}^1{n}_0→{}^{83}{G}_{32}+{}^A{X}_Z$

Equate the above equation for the value of molar mass A.

$$\begin{gathered} 235+1=83+A \\ 236=83+A \\ A=236-83 \\ A=253 \end{gathered}$$

Equate the equation for the value of atomic number Z.

$$\begin{gathered} 92+0=32+Z \\ Z=92-32 \\ Z=60 \end{gathered}$$

According to the molar number and atomic number the other fragment of ${}^{235}U$is neodymium and represents as ${}^{153}N{d}_{60}$.

Hence the other fragment is ${}^{153}Nd$.

The expression for the total energy is given by,

$Q=K{E}_{Ge}+K{E}_{Nd}$ …… (1)

The energy of the germanium is given by,

role="math" localid="1661940308813" $K{E}_{Ge}=\frac{P_{Ge}^2}{2M_{Ge}}$ …… (2)

The energy of the neodymium is given by,

$K{E}_{Nd}=\frac{P_{Nd}^2}{2M_{Nd}}$ …… (3)

By using the conservation linear momentum.

$P_{Ge}^2=P_{Nd}^2$

Substitute $P_{Nd}^2$for $P_{Ge}^2$into equation (2).

$$\begin{gathered} K{E}_{Ge}=\frac{P_{Nd}^2}{2M_{Ge}} \\ {P}_{Nd}^2=2M_{Ge}K{E}_{Ge} \end{gathered}$$

Substitute $2M_{Ge}K{E}_{Ge}$for $P_{Nd}^2$into equation (3).

$$\begin{gathered} K{E}_{Nd}=\frac{2M_{Ge}K{E}_{Ge}}{2M_{Nd}} \\ K{E}_{Nd}=\frac{M_{Ge}}{M_{Nd}}K{E}_{Ge} \end{gathered}$$

…… (4)

Substitute $\frac{M_{Ge}}{M_{Nd}}K{E}_{Ge}$for$K{E}_{Nd}$into equation (1).

$$\begin{gathered} Q=K{E}_{Ge}+\frac{M_{Ge}}{M_{Nd}}K{E}_{Ge} \\ =K{E}_{Ge}\left(1+\frac{M_{Ge}}{M_{Nd}}\right) \\ K{E}_{Ge}=\frac{Q}{1+\frac{M_{Ge}}{M_{Nd}}} \\ K{E}_{Ge}=\frac{Q}{M_{Nd}+M_{Ge}} \end{gathered}$$

…… (5)

Calculate the energy for the germanium.

Substitute 83g/mol for $M_{Ge}$,170mEv for Q and 153g/mol for$M_{Nd}$into equation(5).

$$\begin{gathered} K{E}_{Ge}=\frac{170×153}{153+83} \\ =\frac{170×153}{236} \\ =110.21\mathrm{MeV} \end{gathered}$$

By rounding down the value of the energy for germanium is 110 MeV.

Hence, the energy of the germanium fragment is 110 MeV .

Calculate the energy for the neodymium.

Substitute 83g/mol for $M_{Ge}$,110mEv for$K{E}_{Ge}$ and 153g/mol for $M_{Nd}$into equation (4).

$$\begin{gathered} K{E}_{Nd}=\frac{83}{153}×110 \\ =0.5424×110 \\ =59.78\mathrm{MeV} \end{gathered}$$

By rounding up the energy for the neodymium us 60 MeV.

Hence, the energy of the neodymium is 60 MeV.

As known that,

$$\begin{gathered} 1\mathrm{g}/\mathrm{mol}=1\mathrm{u} \\ =1.66×{10}^{-27}\mathrm{kg} \\ 1\mathrm{MeV}=1.602×{10}^{-13}\frac{\mathrm{kg}.{\mathrm{m}}^2}{{\mathrm{s}}^2} \end{gathered}$$

The speed of the fragment germanium is given by,

$v_{Ge}=\sqrt{\frac{2K{E}_{Ge}}{M_{Ge}}}$

Substitute 83g/mol for $M_{Ge}$,and 110 MeV for $K{E}_{Ge}$into above equation.

$\begin{array}{r}{v}_{Ge}=\sqrt{\frac{2×110×1.602×{10}^{−13}}{83×1.66×{10}^{−27}}}\\ =\sqrt{\frac{352.44×{10}^{−13}}{137.78×{10}^{−27}}}\\ =\sqrt{2.557×{10}^{14}}\\ =1.6×{10}^7\mathrm{m}/\mathrm{s}\end{array}$

Hence, the speed of the germanium is $1.6×{10}^7\mathrm{m}/\mathrm{s}$.

The speed of the fragment neodymium is given by,

$v_{Nd}=\sqrt{\frac{2K{E}_{Nd}}{M_{Nd}}}$

Substitute 153 g/mol for $M_{Nd}$,and 60 MeV for $K{E}_{Nd}$into above equation.

$\begin{array}{r}{v}_{Ge}=\sqrt{\frac{2×60×1.602×{10}^{−13}}{153×1.66×{10}^{−27}}}\\ =\sqrt{\frac{192.24×{10}^{−13}}{253.98×{10}^{−27}}}\\ =\sqrt{0.7569×{10}^{14}}\\ =8.7×{10}^6\mathrm{m}/\mathrm{s}\end{array}$

Hence, the speed of the neodymium is $8.7×{10}^6\mathrm{m}/\mathrm{s}$.