Chapter 45, Problem 72AP

(a)

To determine

The total numbers of fissions have occurred up to and including the $n^{th}$ generation.

Answer to Problem 72AP

The total number of fissions have occurred up to and including the $n^{th}$ generation is $N=N_0\frac{K^{n+1}-1}{K-1}$.

Explanation of Solution

In this case, the number of fissions occurring in the zeroth generation is $N_0$.

The number of fissions occurring in the first generation is $N_{0}K$.

The number of fissions occurring in the second generation is $N_0{K}^2$.

The number of fissions occurring in the $n^{th}$ generation is $N_0{K}^n$.

Write the expression for the total number of fissions occurred up to $n^{th}$ generation.

    $N=N_0+N_{0}K+N_0{K}^2+\mathrm{.......}+N_0{K}^n$                             (I)

Here, $N$ is the total number of fissions occurred up to $n^{th}$ generation, $N_{\text{0}}$ is the number of fissions occurring in the zeroth generation and $K$ is the reproduction constant.

Rewrite the above expression,

    $N=N_0\left(1+K+K^2+\mathrm{...}+K^n\right)$                                      (II)

Write the expression of algebra for the factoring of the difference of two squares.

    $a^2-1=\left(a+1\right)\left(a-1\right)$                                                   (III)

Here, $a$ is the variable.

Write the expression of algebra for the factoring of the difference of two cubes.

    $a^3-1=\left(a^2+a+1\right)\left(a-1\right)$                                             (IV)

Write the expression of algebra for the generalized to the difference of two quantities to any power.

    $a^{n+1}-1=\left(a^n+a^{n-1}+\mathrm{....}+a^2+a+1\right)\left(a-1\right)$                               (V)

Conclusion:

Substitute $K$ for $a$ in (V),

    $K^{n+1}-1=\left(K^n+K^{n-1}+\mathrm{....}+K^2+K+1\right)\left(K-1\right)$

    $\frac{K^{n+1}-1}{K-1}=K^n+K^{n-1}+\mathrm{....}+K^2+K+1$

Substitute $\frac{K^{n+1}-1}{K-1}$ for $K^n+K^{n-1}+\mathrm{....}+K^2+K+1$ in (II),

    $N=N_0\frac{K^{n+1}-1}{K-1}$

Therefore, the total number of fissions have occurred up to and including the $n^{th}$ generation is $N=N_0\frac{K^{n+1}-1}{K-1}$.

(b)

To determine

The uranium in the weapon takes the time to complete the fission after its zeroth generation.

Answer to Problem 72AP

The uranium in the weapon takes the time to complete the fission after its zeroth generation is $1.00\mu \text{s}$.

Explanation of Solution

Write the expression for the total number of fissions have occurred up to and including the $n^{th}$ generation.

    $N=\frac{m}{A}$                                                                                    (VI)

Here, $N$ is the total number of fissions occurred up to $n^{th}$ generation, $m$ is the mass of the uranium and $A$ is the atomic mass.

From part (a),

Write the expression for the total number of fissions have occurred up to and including the $n^{th}$ generation.

    $N=N_0\frac{K^{n+1}-1}{K-1}$                                                                                     (VII)

Here, $N$ is the total number of fissions occurred up to $n^{th}$ generation, $N_{\text{0}}$ is the number of fissions occurring in the zeroth generation, $n$ is the number of fission generation and $K$ is the reproduction constant.

Rewrite the above expression for $n$ .

    $\begin{array}{c}N\left(K-1\right)=N_0{K}^{n+1}-1\\ \frac{N}{N_0}\left(K-1\right)+1=K^n\left(K\right)\\ K^n=\left(\frac{N}{N_0}\left(K-1\right)+1\right)/K\\ n\ln K=\ln \left(\frac{N\left(K-1\right)}{N_0}+1\right)-\ln K\end{array}$

    $n=\frac{\ln \left(N\left(K-1\right)/N_0+1\right)}{K}-1$                                                  (VIII)

Write the expression for the uranium in the weapon takes the time to complete the fission after its zeroth generation.

    $t=n\Delta t$                                                                                       (IX)

Here, $t$ the uranium in the weapon takes the time to complete the fission after its zeroth generation and $\Delta t$ is the average time interval between one fission generation and next fission generation.

Conclusion:

Substitute $5.50\text{kg}$ for $m$ and $235\text{u}$ for $A$ a in (VI) to find $N$.

    $\begin{array}{c}N=\left(5.50\text{kg}\right)\left(\frac{1\text{atom}}{235\text{u}}\right)\left(\frac{1\text{u}}{1.66\times {10}^{-27}\text{kg}}\right)\\ =1.41\times {10}^{25}\text{nuclei}\end{array}$

Substitute $1.41\times {10}^{25}\text{nuclei}$ for $N$, ${10}^{20}$ for $N_0$ and $0.1$ for $K$ in (VIII) to find $n$.

    $\begin{array}{c}n=\frac{\ln \left(\left(1.41\times {10}^{25}\right)\left(0.1\right)/{10}^{20}+1\right)}{\ln 1.1}-1\\ =99.2\end{array}$

Substitute $100$ for n and $10\times {10}^{-9}\text{s}$ for $\Delta t$ in (IX) to find $t$ .

    $\begin{array}{c}t=100\left(10\times {10}^{-9}\text{s}\right)\\ =1.00\times {10}^{-6}\text{s}\\ \text{=1}\text{.00}\mu \text{s}\end{array}$

Therefore, the uranium in the weapon takes the time to complete the fission after its zeroth generation is $1.00\mu \text{s}$.

(c)

To determine

The speed of sound in the uranium.

Answer to Problem 72AP

The speed of sound in the uranium is $2.83\text{km/s}$.

Explanation of Solution

Write the expression for the speed of sound in the uranium.

    $v=\sqrt{\frac{B}{\rho }}$                                                                  (X)

Here, $v$ is the speed of sound in the uranium, $B$ is the bulk modulus of the uranium and $\rho$ is density of the uranium.

Conclusion:

Substitute $150\times {10}^9{\text{N/m}}^{\text{2}}$ for $B$ and $18.7\times {10}^3{\text{kg/m}}^3$ for $\rho$ in (X) to find $v$.

    $\begin{array}{c}v=\sqrt{\frac{150\times {10}^9{\text{N/m}}^{\text{2}}}{18.7\times {10}^3{\text{kg/m}}^3}}\\ =2.83\times {10}^3\text{m/s}\\ \text{=2}\text{.83}\text{km/s}\end{array}$

Therefore, the speed of sound in the uranium is $2.83\text{km/s}$.

(d)

To determine

The time interval for a compressional wave to cross the radius of a $5.50\text{kg}$ sphere of uranium.

Answer to Problem 72AP

The time interval for a compressional wave to cross the radius of a $5.50\text{kg}$ sphere of uranium is $14.6\mu \text{s}$.

Explanation of Solution

Write the expression for the volume of the uranium.

    $V=\frac{4}{3}\pi r^3$                                                                (XI)

Here, $V$ is the volume of the uranium and  $r$ is the radius of the uranium.

Write the expression for the volume of the uranium in mass density.

    $V=\frac{m}{\rho }$                                                                      (XII)

Here, $V$ is the volume of the uranium and  $m$ is the mass of the uranium and $\rho$ is density of the uranium

Equating expression (XI) and (XII),

    $\frac{4}{3}\pi r^3=\frac{m}{\rho }$

    $r={\left(\frac{3m}{4\pi \rho }\right)}^{1/3}$                                                          (XIII)

Write the expression for the time interval for a compressional wave.

    $\Delta t_d=\frac{r}{v}$                                                                (XIV)

Here, $\Delta t_d$ is the time interval for a compressional wave and $v$ is the speed of the sound in uranium.

Conclusion:

Substitute $5.50\text{kg}$ for $m$ and $18.7\times {10}^3{\text{kg/m}}^3$ for $\rho$ in (XIII) to find $r$.

    $\begin{array}{c}r={\left(\frac{3\left(5.50\text{kg}\right)}{4\pi \left(18.7\times {10}^3{\text{kg/m}}^3\right)}\right)}^{1/3}\\ =4.13\times {10}^{-2}\text{m}\end{array}$

Substitute $4.13\times {10}^{-2}\text{m}$ for $r$ and $2.83\text{km/s}$ for $v$ in (XIV) to find $\Delta t_d$.

    $\begin{array}{c}\Delta t_d=\frac{4.13\times {10}^{-2}\text{m}}{2.83\times {10}^3\text{m/s}}\\ =1.46\times {10}^{-5}\text{s}\\ =14.6\mu \text{s}\end{array}$

Therefore, the time interval for a compressional wave to cross the radius of a $5.50\text{kg}$ sphere of uranium is $14.6\mu \text{s}$.

(e)

To determine

Does the entire bomb release the explosive energy of all its uranium.

Does the explosive energy of the uranium release in equivalent tons of TNT.

Answer to Problem 72AP

Yes, the entire bomb release the explosive energy of all its uranium

The explosive energy of the uranium is release $108\text{kilotons}$ of TNT.

Explanation of Solution

Write the expression for the ratio of the energy release between the uranium bomb and one ton of TNT.

    $NE=E_1$                                                                  (XV)

Here, $N$ is the total number of fissions occurred up to $n^{th}$ generation, $E$ is the energy of the uranium bomb and $E_1$ is energy of TNT bomb.

Conclusion:

Substitute $1.41\times {10}^{25}\text{nuclei}$ for $N$, $200\times {10}^6\text{eV}$ for $E$ and $4.51\times {10}^{14}\text{J}$ for $E_1$ in (XV).

    $\left(1.41\times {10}^{25}\text{nuclei}\right)\left(\frac{200\times {10}^6\text{eV}}{1\text{nucleus}}\right)\left(\frac{1.60\times {10}^{-19}\text{J}}{1\text{eV}}\right)=\left(4.51\times {10}^{14}\text{J}\right)\left(\frac{1\text{ton}\text{of TNT}}{4.20\times {10}^9\text{J}}\right)$

    $\begin{array}{l}=1.08\times {10}^5\text{ton}\text{TNT}\\ \text{=108}\text{kilotons of TNT}\end{array}$

Therefore, the entire bomb release the explosive energy of all its uranium

The explosive energy of the uranium is release $108\text{kilotons}$ of TNT.