Answer
See answers.
Work Step by Step
a. We find the wavelength using equation 27–8.
$$\lambda=\frac{h}{p}=\frac{h}{mv}$$
$$\lambda=\frac{6.626\times10^{-34}J \cdot s }{(0.012kg)(150m/s)}=3.7\times10^{-34}m$$
b. The uncertainty in the bullet’s position is given. Use equation 28–1 to find the uncertainty in the momentum.
$$\Delta p \geq \frac{\hbar}{\Delta x}$$
$$=\frac{(1.055\times10^{-34}J\cdot s)}{0.0060m}= 1.8\times10^{-32}kg \cdot m/s$$
