## College Physics (4th Edition)

We can find the wavelength of a photon with an energy of $0.100~keV$: $E = \frac{hc}{\lambda_p}$ $\lambda_p = \frac{hc}{E}$ $\lambda_p = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(0.100\times 10^3~eV)(1.6\times 10^{-19}~J/eV)}$ $\lambda_p = 1.2424\times 10^{-8}~m$ We can find the wavelength of the electron: $\lambda_e = \frac{h}{p_e}$ $\lambda_e = \frac{h}{\sqrt{2~m_e~E_e}}$ $\lambda_e = \frac{6.626\times 10^{-34}~J~s}{\sqrt{(2)(9.1\times 10^{-31}~kg)(100~eV)(1.6\times 10^{-19}~J/eV)}}$ $\lambda_e = 1.228\times 10^{-10}~m$ We can find the ratio of the wavelengths: $\frac{\lambda_p}{\lambda_e} = \frac{1.2424\times 10^{-8}~m}{1.228\times 10^{-10}~m} = 100$ The ratio of the wavelength of the photon to the wavelength of the electron is 100.