Chapter 28, Problem 40P

(a)

To determine

The minimum average distance by which the target was missed.

Answer to Problem 40P

The minimum average miss distance is ${\left(\frac{2\hslash }{m}\right)}^{1/2}{\left(\frac{2H}{g}\right)}^{1/4}$.

Explanation of Solution

The initial height of the pellet is $H$, its mass is $m$ and let the target be the origin

Write the given expression for deviation from the target

  $\Delta x_f=\left(\Delta x_i\right)+\left(\Delta v_{x}t\right)$                                                                                                    (I)

Here, $\Delta x_f$ is the uncertainty in position, $\Delta x_f$ is the uncertainty in initial position, $\Delta v_x$ is the uncertainty in velocity and $t$ is the time taken.

Write the uncertainty relation.

  $\Delta x_i\left(m\Delta v_x\right)=\frac{\hslash }{2}$

Here, $\Delta x_f$ is the uncertainty in position, $\Delta v_x$ is the uncertainty in velocity, $m$ is the mass and $\hslash$ is the reduced Planck’s constant.

Substitute (I) and rearrange

  $\Delta v_x=\frac{\hslash }{2\left(m\Delta x_i\right)}$                                                                                                          (II)

Write the expression for change in velocity using equation of motion

    $s=ut+\frac{1}{2}a{t}^2$                                                                                                      (III)

Here, $u$ is initial velocity, $a$ is the acceleration, $s$ is the distance travelled.

Substitute $0$ for $u$, $g$ for $a$ and $H$ for $s$ in (III) and rearrange for $t$

Write the expression for time taken to reach distance $H$

  $t=\sqrt{\frac{2H}{g}}$                                                                                                             (IV)

Substitute (IV) and (II) in (I)

  $\Delta x_f=\left(\Delta x_i\right)+\frac{\hslash }{2m\Delta x_i}\sqrt{\frac{2H}{g}}$                                                                                       (V)

Find the uncertainty in initial position by differentiating the above equation with respect to $\Delta x_i$

  $\begin{array}{c}\frac{d\left(\Delta x_f\right)}{d\left(\Delta x_i\right)}=0\\ 1-\frac{\hslash }{2m{\left(\Delta x_i\right)}^2}\sqrt{\frac{2H}{g}}=0\end{array}$

Since the second derivative of equation (V) is positive. The above found $\Delta x_i$ is the minimum.

Rearrange for $\Delta x_i$

  $\Delta x_i={\left(\frac{\hslash }{2m}\right)}^{1/2}{\left(\frac{2H}{g}\right)}^{1/4}$                                                                                            (V)

Conclusion

Substitute $\Delta x_i$ in (V) to find $\Delta x_f$

  $\begin{array}{c}\Delta x_f={\left(\frac{\hslash }{2m}\right)}^{1/2}{\left(\frac{2H}{g}\right)}^{1/4}+\frac{\hslash }{2m}\sqrt{\frac{2H}{g}}{\left({\left(\frac{\hslash }{2m}\right)}^{1/2}{\left(\frac{2H}{g}\right)}^{1/4}\right)}^{-1}\\ =2\left({\left(\frac{\hslash }{2m}\right)}^{1/2}{\left(\frac{2H}{g}\right)}^{1/4}\right)\\ ={\left(\frac{2\hslash }{m}\right)}^{1/2}{\left(\frac{2H}{g}\right)}^{1/4}\end{array}$

Thus, the minimum average miss distance is ${\left(\frac{2\hslash }{m}\right)}^{1/2}{\left(\frac{2H}{g}\right)}^{1/4}$.

(b)

To determine

Find the average miss distance given the mass and height.

Answer to Problem 40P

The average miss distance is $5.19\times {10}^{-16}\text{ m}$.

Explanation of Solution

The mass of the pellet is $0.500\text{g}$ and the height from which it is dropped is $2.00\text{m}$.

Substitute $0.500\text{g}$ for $m$, $2.00\text{m}$ for $H$, $9.8{\text{ms}}^{-2}$ for $g$ and $1.054\times {10}^{-34}\text{ Js}$ for $\hslash$ in (V) to find $\Delta x_f$

    $\begin{array}{c}\Delta x_f={\left(\frac{2\times 1.054\times {10}^{-34}{\text{ kgm}}^2{\text{s}}^{-1}}{0.500\text{g}}\right)}^{1/2}{\left(\frac{2\times 2.00\text{m}}{9.8{\text{ms}}^{-2}}\right)}^{1/4}\\ ={\left(\frac{2\times 1.054\times {10}^{-34}{\text{ kgm}}^2{\text{s}}^{-1}}{0.500\times {10}^{-3}\text{kg}}\right)}^{1/2}{\left(\frac{2\times 2.00\text{m}}{9.8{\text{ms}}^{-2}}\right)}^{1/4}\\ =5.19\times {10}^{-16}\text{ m}\end{array}$

Thus, the average miss distance is $5.19\times {10}^{-16}\text{ m}$.