Answer
See answers.
Work Step by Step
a. Calculate the momentum using equation 27–8.
$$\lambda=\frac{h}{p}$$
$$p=\frac{h}{\lambda}=\frac{6.626\times10^{-34}J \cdot s }{4.5\times10^{-10}m}$$
$$=1.5\times10^{-24}kg \cdot m/s$$
b. Assume that the electron is nonrelativistic. Find the speed using equation 27–8.
$$\lambda=\frac{h}{p}=\frac{h}{mv}$$
$$v=\frac{h}{m\lambda}=\frac{6.626\times10^{-34}J \cdot s }{(9.11\times10^{-31}kg)(4.5\times10^{-10}m)}$$
$$=1.6\times10^{6}m/s$$
Our assumption is valid because v is less than 1 percent of c.
c. First, calculate the kinetic energy of the electron.
$$KE=\frac{1}{2}mv^2=\frac{1}{2}(9.11\times10^{-31}kg)( 1.616\times10^{6}m/s)^2 =1.19\times10^{-18}J\approx 7.4 eV$$
To give an electron 7.4 eV of kinetic energy, the electron at rest must be accelerated through a potential difference of 7.4 volts.