## University Physics with Modern Physics (14th Edition)

a. $E=\frac{hc}{\lambda}=\frac{(4.136\times10^{-15}eV)(3.00\times10^8 m/s)}{0.20\times10^{-9}m}=6.2keV$ b. $E=\frac{p^2}{2m}=(\frac{h}{\lambda})^2 \frac{1}{2m}=\frac{h^2}{2m \lambda^2}$ Evaluate using the electron’s mass to find 38 eV. c. Evaluate the expression using the alpha particle’s mass to find an energy of $5.2\times10^{-3}eV$. For a given wavelength, a photon has much more energy than an electron. The electron has more energy than an alpha particle.