Chapter 14, Problem 60P

(a)

To determine

Prove that the number of fissions is given by the formula $N=N_0\left(\frac{K^{n+1}-1}{K-1}\right)$.

Answer to Problem 60P

It is proved that the number of fissions is given by $N=N_0\left(\frac{K^{n+1}-1}{K-1}\right)$.

Explanation of Solution

The number of fissions in 1^st, 2^nd, 3^rd …. nth number is according to the series given below.

   $N_0,N_{0}K,N_0{K}^2\mathrm{....}{N}_0{K}^n$

Write the equation to find the total number of fissions.

    $\begin{array}{c}N=N_0+N_{0}K+N_0{K}^2+N_0{K}^n\\ =N_0\left(1+K+K^2+K^n\right)\\ =N=N_0\left(\frac{K^{n+1}-1}{K-1}\right)\end{array}$

Conclusion:

Thus, it is proved that the number of fissions is given by $N=N_0\left(\frac{K^{n+1}-1}{K-1}\right)$.

(b)

To determine

Time needed for the complete fission of uranium bomb.

Answer to Problem 60P

Time required will be $1.00\times {10}^{-6}\text{s}$.

Explanation of Solution

Write the equation to find the number of uranium nuclei.

    $N=\frac{m}{m_p{n}_p}$        (I)

Here, $N$ the number of uranium nuclei, $m$ the mass of uranium, $n_p$ the number of protons in uranium nuclei and $m_p$ the mass of proton.

Write the equation to find the number of fissions given in part (a).

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

Write the equation to find the total time to complete fission from the time taken for the fission of one nucleus.

    $\Delta t=n\Delta t_1$        (III)

Here $\Delta t$ the total time and $\Delta t_1$ is the time taken for the fission of one nucleus.

Conclusion:

Substitute $5.50\text{ kg}$ $m$, 235 for $n_p$ $\text{1}\text{.66}\times {\text{10}}^{-27}\text{ kg}$ $m_p$ equation (I).

    $\begin{array}{c}N=\frac{5.50\text{ kg}}{\left(\text{1}\text{.66}\times {\text{10}}^{-27}\text{ kg}\right)\left(235\right)}\\ =1.41\times {10}^{25}\text{nuclie}\end{array}$

Substitute $1.41\times {10}^{25}\text{nuclie}$ for $N$, 1.10 for $K$ and $1.00\times {10}^{20}\text{fissions}$ $N_0$ equation (II).

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

Substitute $\begin{array}{c}n=\frac{\ln \left(\frac{1.41\times {10}^{25}\text{nuclie}\left(1.10-1\right)}{1.00\times {10}^{20}\text{fissions}}+1\right)}{\ln 1.10}-1\\ =99.2\end{array}$

So, the number of generations can be taken as 100.

Substitute 100 for $n$ and 10.0 ns for $\Delta t_1$ equation (III).

    $\begin{array}{c}\Delta t=\left(100\right)\left(10.0\text{ns}\left(\frac{{10}^{-9}\text{s}}{1\text{ns}}\right)\right)\\ =1.00\times {10}^{-6}\text{ s}\end{array}$

Thus, time required will be $1.00\times {10}^{-6}\text{s}$.

(c)

To determine

Speed of sound in uranium.

Answer to Problem 60P

Speed is $2.83\times {10}^{-6}\text{m}/\text{s}$.

Explanation of Solution

Write the equation to find the speed of sound through a medium.

    $v=\sqrt{\frac{B}{\rho }}$

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

Conclusion:

Substitute $150\times {10}^9\text{N}/{\text{m}}^{\text{2}}$ $B$ $18.7\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ $\rho$ the above equation.

    $\begin{array}{c}v=\sqrt{\frac{150\times {10}^9\text{N}/{\text{m}}^{\text{2}}}{18.7\times {10}^3\text{kg}/{\text{m}}^{\text{3}}}}\\ =2.83\times {10}^{-6}\text{m}/\text{s}\end{array}$

Thus, the speed is $2.83\times {10}^{-6}\text{m}/\text{s}$.

(d)

To determine

Time taken by compressional wave to cross the uranium sphere.

Answer to Problem 60P

Time taken is $1.46\times {10}^{-5}\text{ s}$.

Explanation of Solution

Write the equation for the volume of uranium sphere.

    $V=\frac{4}{3}\pi r^3$

Here, $r$ is the radius of uranium sphere.

Express $V$ terms of mass and density.

    $V=\frac{m}{\rho }$

Here, $m$ the mass and $\rho$ the density.

Equate the right-hand sides of above two equations.

    $\begin{array}{c}\frac{4}{3}\pi r^3=\frac{m}{\rho }\\ r={\left(\frac{3m}{4\pi \rho }\right)}^{1/3}\end{array}$

Write the equation to find the time taken by compressional wave to cross the uranium sphere.

    $\Delta t_d=\frac{r}{v}$

Conclusion:

Substitute $5.5\text{kg}$ $m$ $18.7\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ $\rho$ the equation for $r$.

    $\begin{array}{c}r={\left(\frac{3\left(5.5\text{kg}\right)}{4\pi \left(18.7\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\right)}\right)}^{1/3}\\ =4.13\times {10}^{-2}\text{m}\end{array}$

Substitute $4.13\times {10}^{-2}\text{m}$ $r$ $\text{2}\text{.83}\times {\text{10}}^{\text{3}}\text{m}/\text{s}$ $v$ the equation for $\Delta t_d$.

    $\begin{array}{c}\Delta t_d=\frac{4.13\times {10}^{-2}\text{m}}{\text{2}\text{.83}\times {\text{10}}^{\text{3}}\text{m}/\text{s}}\\ =1.46\times {10}^{-5}\text{s}\end{array}$

Thus, the time taken is $1.46\times {10}^{-5}\text{ s}$.

(d)

To determine

Check whether the bomb explained in part (b) can give energy by using all of its uranium and if possible, calculate the amount of energy in TNT.

Answer to Problem 60P

The entire uranium can undergo fission and it releases 107 kilotons of TNT.

Explanation of Solution

Time taken by the compressional wave to cross the entire uranium sphere is $14.6\text{μs}$ and the time taken for complete fission is only $1.0\text{μs}$. Since the time taken by the wave to cross the sphere is greater than the fission time, bomb can utilize all of its uranium.

Write the equation to find the energy released by the bomb.

    $E=n{E}_{\text{per}\text{fission}}$

Here, $n$ is the number of nuclei and the $E_{\text{per}\text{fission}}$ the energy produced per fission.

Conclusion:

Substitute $1.41\times {10}^{25}\text{nuclei}$ $n$ $\text{200}\text{MeV}$ $E_{\text{per}\text{fission}}$ the above equation.

    $\begin{array}{c}E=\left(1.41\times {10}^{25}\text{nuclei}\right)\left(\text{200}\text{MeV}\left(\frac{1.6\times {10}^{-19}\text{ J}}{\text{1 eV}}\right)\right)\\ =4.51\times {10}^{14}\text{ J}\\ =4.51\times {10}^{14}\text{ J}\left(\frac{\text{1 ton TNT}}{\text{4}\text{.2}\times {\text{10}}^{\text{9}}\text{ J}}\right)\\ =1.07\times {10}^5\text{ ton TNT}\left(\frac{\text{1}\text{kiloton}}{{\text{10}}^{\text{3}}\text{ ton}}\right)\\ =107\text{kiloton}\text{TNT}\end{array}$

Thus, the entire uranium can undergo fission and it releases 107 kilotons of TNT.