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 electron’s momentum.
$$\Delta p=m\Delta v \geq \frac{\hbar}{\Delta x}$$
From that, find the uncertainty in its speed.
$$\Delta v \geq \frac{\hbar}{m\Delta x}$$
$$=\frac{(1.055\times10^{-34}J\cdot s)}{(9.11\times10^{-31}kg)(15\times10^{-9}m)}= 7.7\times10^3m/s$$
The speed is nonrelativistic. Calculate the electron's minimum kinetic energy.
$$KE=\frac{1}{2}mv^2=\frac{1}{2}(9.11\times10^{-31}kg)(7720m/s)^2$$
$$=2.7\times10^{-23}J$$
This is about 0.00017 eV.