College Physics (4th Edition)

The change in velocity is $0.078~m/s$
We can use the kinetic energy $E$ to write an expression for the speed of a particle: $E = \frac{1}{2}mv^2$ $v = \sqrt{\frac{2~E}{m}}$ We can find the speed of the particle when it absorbs a photon with a wavelength of $660~nm$: $v = \sqrt{\frac{2~E}{m}}$ $v = \sqrt{\frac{2~(\frac{hc}{\lambda})}{m}}$ $v = \sqrt{\frac{2~h~c}{m~\lambda}}$ $v = \sqrt{\frac{(2)(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{(1.0\times 10^{-16}~kg)(660\times 10^{-9}~m)}}$ $v = 0.078~m/s$ The change in velocity is $0.078~m/s$