Chapter 28, Problem 37E

(a)

To determine

The energy difference between $n=1$ and $n=2$ states if the length of the box is $3\text{fm}$.

Answer to Problem 37E

The energy difference is $68.4\text{MeV}$.

Explanation of Solution

Write the expression for the energy of the $n^{th}$ state of a particle confined in a box.

    $E_n=\frac{n^2{h}^2}{8m{a}^2}$        (I)

Here, $E_n$ is the energy in the $n^{th}$ state, $h$ is Planck’s constant, $m$ is the mass of the particle and $a$ is the dimension of the box.

Conclusion:

Substitute $2$ for $n$, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $1.67\times {10}^{-27}\text{kg}$ for $m$ and $3\text{fm}$ for $a$ in equation (I).

    $\begin{array}{c}{E}_2=\frac{2^2{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{8\left(1.67\times {10}^{-27}\text{kg}\right){\left(3\text{fm}\right)}^2}\frac{{\left(1\text{fm}\right)}^2}{{\left(10{}^{-15}\text{m}\right)}^2}\\ =1.46\times {10}^{-11}\text{J}\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =9.12\times {10}^7\text{eV}\\ \text{=}\text{91}\text{.2}\text{MeV}\end{array}$

Substitute $1$ for $n$, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $1.67\times {10}^{-27}\text{kg}$ for $m$ and $3\text{fm}$ for $a$ in expression (I).

  $\begin{array}{c}{E}_1=\frac{1^2{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{8\left(1.67\times {10}^{-27}\text{kg}\right){\left(3\text{fm}\right)}^2}\frac{{\left(1\text{fm}\right)}^2}{{\left(10{}^{-15}\text{m}\right)}^2}\\ =3.65\times {10}^{-12}\text{J}\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =2.28\times {10}^7\text{eV}\\ \text{=}22.8\text{MeV}\end{array}$

Evaluate the difference between $E_2$ and $E_1$.

    $\begin{array}{c}{E}_2-E_1=\left(91.2-22.8\right)\text{MeV}\\ \text{=}\text{68}\text{.4}\text{MeV}\end{array}$

Thus, the energy difference is $68.4\text{MeV}$.

(b)

To determine

The energy difference between $n=1$ and $n=2$ states if the length of the box is $8\text{fm}$.

Answer to Problem 37E

The energy difference is $9.59\text{MeV}$.

Explanation of Solution

Write the expression for the energy of the $n^{th}$ state of a particle confined in a box.

    $E_n=\frac{n^2{h}^2}{8m{a}^2}$

Here, $E_n$ is the energy in the $n^{th}$ state, $h$ is Planck’s constant, $m$ is the mass of the particle and $a$ is the dimension of the box.

Conclusion:

Substitute $2$ for $n$, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $1.67\times {10}^{-27}\text{kg}$ for $m$ and $8\text{fm}$ for $a$ in expression (I).

    $\begin{array}{c}{E}_2=\frac{2^2{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{8\left(1.67\times {10}^{-27}\text{kg}\right){\left(\text{8fm}\right)}^2}\frac{{\left(1\text{fm}\right)}^2}{{\left(10{}^{-15}\text{m}\right)}^2}\\ =2.05\times {10}^{-12}\text{J}\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =1.28\times {10}^7\text{eV}\\ \text{=}12.8\text{MeV}\end{array}$

Substitute $1$ for $n$, $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$, $1.67\times {10}^{-27}\text{kg}$ for $m$ and $8\text{fm}$ for $a$ in expression (I).

  $\begin{array}{c}{E}_1=\frac{1^2{\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}^2}{8\left(1.67\times {10}^{-27}\text{kg}\right){\left(\text{8fm}\right)}^2}\frac{{\left(1\text{fm}\right)}^2}{{\left(10{}^{-15}\text{m}\right)}^2}\\ =5.13\times {10}^{-13}\text{J}\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =3.21\times {10}^6\text{eV}\\ \text{=}3.21\text{MeV}\end{array}$

Evaluate the difference between $E_2$ and $E_1$.

    $\begin{array}{c}{E}_2-E_1=\left(12.8-3.21\right)\text{MeV}\\ \text{=}9.59\text{MeV}\end{array}$

Thus, the energy difference is $9.59\text{MeV}$.