Chapter 27, Problem 46AP

(a)

To determine

Johnny’s de Broglie wavelength at the moment.

Answer to Problem 46AP

Johnny’s de Broglie wavelength at the moment is $2.82\times {10}^{-37}\text{m}$.

Explanation of Solution

Johnny’s speed just before impact is,

$\begin{array}{l}{v}^2=v_0^2+2a_y\left(\Delta y\right)\\ v=\sqrt{v_0^2+2a_y\left(\Delta y\right)}\end{array}$

• • $v_0$ is the initial velocity
  • $a_y$ is the vertical acceleration
  • $\Delta y$ is the vertical displacement

Substitute $0$ for $v_0$, $9.80\text{m/s}$ for $a_y$ and $50.0\text{m}$ for $\Delta y$.

$\begin{array}{c}v=\sqrt{0+2\left(9.80\text{m/s}\right)\left(50.0\text{m}\right)}\\ =31.3\text{m/s}\end{array}$

The equation for de Broglie wavelength is,

$\lambda =\frac{h}{mv}$

• • $h$ is the Plank’s constant
  • $m$ is the mass
  • $v$ is the velocity

Substitute $6.63\times {10}^{-34}\text{J}\text{.s}$ for $h$, $75.0\text{kg}$ for $m$ and $31.3\text{m/s}$ for $v$.

$\begin{array}{c}\lambda =\frac{6.63\times {10}^{-34}\text{J}\text{.s}}{\left(75.0\text{kg}\right)\left(31.3\text{m/s}\right)}\\ =2.82\times {10}^{-37}\text{m}\end{array}$

Conclusion:

Thus, Johnny’s de Broglie wavelength at the moment is $2.82\times {10}^{-37}\text{m}$.

(b)

To determine

The uncertainty of the kinetic energy measurement.

Answer to Problem 46AP

The uncertainty of the kinetic energy measurement is $1.06\times {10}^{-32}\text{J}$.

Explanation of Solution

The equation for energy uncertainty is,

$\Delta E\ge \frac{h}{4\pi \left(\Delta t\right)}$

Substitute $6.63\times {10}^{-34}\text{J}\text{.s}$ for $h$ and $5.00\text{ms}$ for $\Delta t$.

$\begin{array}{c}\Delta E=\frac{6.63\times {10}^{-34}\text{J}\text{.s}}{4\pi \left(5.00\text{ms}\right)\left(\frac{{10}^{-3}\text{s}}{1\text{ms}}\right)}\\ =1.06\times {10}^{-32}\text{J}\end{array}$

Conclusion:

Thus, the uncertainty of the kinetic energy measurement is $1.06\times {10}^{-32}\text{J}$.

(c)

To determine

The percentage error caused by the uncertainty.

Answer to Problem 46AP

The percentage error caused by the uncertainty is $2.88\times {10}^{-35}\%$.

Explanation of Solution

The percentage error is,

$\%\text{error}=\frac{\Delta E}{mg\left|\Delta y\right|}\left(100\%\right)$

Substitute $1.06\times {10}^{-32}\text{J}$ for $\Delta E$, $75.0\text{kg}$ for $m$, $9.80{\text{m/s}}^2$ for $g$ and $50.0\text{m}$ for $\Delta y$.

$\begin{array}{c}\%\text{error}=\frac{1.06\times {10}^{-32}\text{J}}{\left(75.0\text{kg}\right)\left(9.80{\text{m/s}}^2\right)\left(50.0\text{m}\right)}\left(100\%\right)\\ =2.88\times {10}^{-35}\%\end{array}$

Conclusion:

Thus, the percentage error caused by the uncertainty is $2.88\times {10}^{-35}\%$