## College Physics (4th Edition)

(a) $\lambda = 1.23\times 10^{-9}~m$ (b) $\lambda = 3.88\times 10^{-11}~m$
(a) We can write an expression for the momentum: $p = \sqrt{2mE}$ We can find the de Broglie wavelength: $\lambda = \frac{h}{p}$ $\lambda = \frac{h}{\sqrt{2mE}}$ $\lambda = \frac{6.626\times 10^{-34}~J~s}{\sqrt{(2)(9.1\times 10^{-31}~kg)(1.0~eV)(1.6\times 10^{-19}~J/eV)}}$ $\lambda = 1.23\times 10^{-9}~m$ (b) We can write an expression for the momentum: $p = \sqrt{2mE}$ We can find the de Broglie wavelength: $\lambda = \frac{h}{p}$ $\lambda = \frac{h}{\sqrt{2mE}}$ $\lambda = \frac{6.626\times 10^{-34}~J~s}{\sqrt{(2)(9.1\times 10^{-31}~kg)(1000~eV)(1.6\times 10^{-19}~J/eV)}}$ $\lambda = 3.88\times 10^{-11}~m$