Answer
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Work Step by Step
The uncertainty in the electron’s position is given. Use equation 28–1 to find the uncertainty in the momentum.
$$\Delta p \geq \frac{\hbar}{\Delta x}$$
$$=\frac{(1.055\times10^{-34}J\cdot s)}{2.0\times10^{-8}m}= 5.3\times10^{-27}kg \cdot m/s$$
Assume that its momentum is this value, and estimate the de Broglie wavelength.
$$\lambda=\frac{h}{p}=\frac{(6.626\times10^{-34}J\cdot s)}{ 5.275\times10^{-27}kg \cdot m/s }= 1.3\times10^{-7}m$$
