Answer
See answers.
Work Step by Step
All 3 energies are much less than the rest energy of an electron. We may use nonrelativistic relationships.
$$KE=\frac{p^2}{2m}$$
$$p=\sqrt{2m(KE)}$$
Calculate the wavelength using equation 27–8.
$$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2m(KE)}}$$
a.
$$\lambda=\frac{6.626\times10^{-34}J \cdot s }{\sqrt{2(9.11\times10^{-31}kg)(10eV)(1.60\times10^{-19}J/eV)}}$$
$$=3.88\times10^{-10}m\approx 4\times10^{-10}m $$
b.
$$\lambda=\frac{6.626\times10^{-34}J \cdot s }{\sqrt{2(9.11\times10^{-31}kg)(100eV)(1.60\times10^{-19}J/eV)}}$$
$$=1.23\times10^{-10}m\approx 1\times10^{-10}m $$
c.
$$\lambda=\frac{6.626\times10^{-34}J \cdot s }{\sqrt{2(9.11\times10^{-31}kg)(1.0\times10^3 eV)(1.60\times10^{-19}J/eV)}}$$
$$=3.88\times10^{-11}m\approx 3.9\times10^{-11}m $$