Chapter 27, Problem 48E

(a)

To determine

To show that the minimum kinetic energy is larger than $1\text{MeV}$.

Answer to Problem 48E

The energy of electron is $3.8125\times {10}^8\text{eV}$ which is greater than $1\text{MeV}$.

Explanation of Solution

Uncertainty principle states that the position and momentum of an atom cannot be measured simultaneously. It gives the uncertainty in position if we have uncertainty in momentum. The equation of uncertainty principle is given as

  $\Delta x\Delta p\ge \hslash$                                                                                                

Here, $\Delta x$ is the change in position, $\Delta p$ is the change in momentum and $\hslash$ is the reduced Planck’s constant.

Rearrange the above expression for $\Delta p$.

    $\Delta p\ge \frac{\hslash }{\Delta x}$                                                                                                

Write the expression for energy.

    $E=\frac{{\left(\Delta p\right)}^2}{2m}$        (I)

Here, $E$ is the energy, $\Delta p$ is change in momentum and $m$ is the mass.

Conclusion:

Substitute $\frac{\hslash }{\Delta x}$ for $\Delta p$ in equation (I).

    $\begin{array}{c}E\ge \frac{{\left(\frac{\hslash }{\Delta x}\right)}^2}{2m}\\ =\frac{{\hslash }^2}{2m{\left(\Delta x\right)}^2}\end{array}$

Substitute $1.055\times {10}^{-34}\text{Js}$ for $\hslash$, $9.1\times {10}^{-31}\text{kg}$ for $m$ and ${10}^{-14}\text{m}$ for $\Delta x$ in above equation.

    $\begin{array}{c}E\ge \frac{{\left(1.055\times {10}^{-34}\text{Js}\right)}^2}{2\left(9.1\times {10}^{-31}\text{kg}\right){\left({10}^{-14}\text{m}\right)}^2}\\ =\frac{{\left(1.055\times {10}^{-34}\text{Js}\right)}^2}{2\left(9.1\times {10}^{-31}\text{kg}\right){\left({10}^{-14}\text{m}\right)}^2}\left(\frac{1{\text{kgm}}^2{\text{s}}^{-2}}{1\text{J}}\right)\\ =6.1\times {10}^{-11}\text{J}\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =3.8125\times {10}^8\text{eV}\end{array}$

Thus, the energy of electron is $3.8125\times {10}^8\text{eV}$ which is greater than $1\text{MeV}$.

(b)

To determine

To see that the energy of neutron is greater than $1\text{MeV}$ or not.

Answer to Problem 48E

The energy of neutron is $2.06\times {10}^5\text{eV}$ which is less than $1\text{MeV}$.

Explanation of Solution

Uncertainty principle states that the position and momentum of an atom cannot be measured simultaneously. It gives the uncertainty in position if we have uncertainty in momentum. The equation of uncertainty principle is given as

  $\Delta x\Delta p\ge \hslash$                                                                                                

Here, $\Delta x$ is the change in position, $\Delta p$ is the change in momentum and $\hslash$ is the reduced Planck’s constant.

Rearrange the above expression for $\Delta p$.

    $\Delta p\ge \frac{\hslash }{\Delta x}$                                                                                                

Write the expression for energy.

    $E=\frac{{\left(\Delta p\right)}^2}{2m}$        (II)

Here, $E$ is the energy, $\Delta p$ is change in momentum and $m$ is the mass.

Conclusion:

Substitute $\frac{\hslash }{\Delta x}$ for $\Delta p$ in equation (II).

    $\begin{array}{c}E\ge \frac{{\left(\frac{\hslash }{\Delta x}\right)}^2}{2m}\\ =\frac{{\hslash }^2}{2m{\left(\Delta x\right)}^2}\end{array}$

Substitute $1.055\times {10}^{-34}\text{Js}$ for $\hslash$, $1.67\times {10}^{-27}\text{kg}$ for $m$ and ${10}^{-14}\text{m}$ for $\Delta x$ in above equation.

    $\begin{array}{c}E\ge \frac{{\left(1.055\times {10}^{-34}\text{Js}\right)}^2}{2\left(1.67\times {10}^{-27}\text{kg}\right){\left({10}^{-14}\text{m}\right)}^2}\\ =\frac{{\left(1.055\times {10}^{-34}\text{Js}\right)}^2}{2\left(1.67\times {10}^{-27}\text{kg}\right){\left({10}^{-14}\text{m}\right)}^2}\left(\frac{1{\text{kgm}}^2{\text{s}}^{-2}}{1\text{J}}\right)\\ =3.3\times {10}^{-14}\text{J}\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ =2.06\times {10}^5\text{eV}\end{array}$

Thus, the energy of neutron is $2.06\times {10}^5\text{eV}$ which is less than $1\text{MeV}$.