91Ó°ÊÓ

We are given the wavelength of the photon as \(660 \mathrm{nm}\). First, let's convert it to meters: \(660 \mathrm{nm} = 660 \times 10^{-9} \mathrm{m}\). Now we can find the energy of the photon using the formula \(E = \frac{hc}{\lambda}\), where \(h = 6.63 \times 10^{-34} \mathrm{Js}\) (Planck's constant) and \(c = 3.0 \times 10^8 \mathrm{m/s}\) (speed of light). Therefore, \(E = \frac{(6.63 \times 10^{-34} \mathrm{Js})(3.0 \times 10^8 \mathrm{m/s})}{(660 \times 10^{-9} \mathrm{m})} = 3.01 \times 10^{-19} \mathrm{J}\).

When the particle absorbs the photon, the momentum of the photon is transferred to the particle. The momentum of the photon can be found using the formula \(p = \frac{E}{c}\), where \(p\) is the momentum, \(E\) is the energy we found in Step 1, and \(c\) is the speed of light. Thus, \(p = \frac{3.01 \times 10^{-19} \mathrm{J}}{3.0 \times 10^8 \mathrm{m/s}} = 1.0 \times 10^{-27} \mathrm{kg \cdot m/s}\). Now we can find the change in the particle's velocity using the conservation of momentum. The momentum of the particle after the collision is the same as the momentum of the photon. So, we have \(\Delta p = m \Delta v\), where \(m = 1.0 \times 10^{-16} \mathrm{kg}\) is the mass of the particle and \(\Delta v\) is the change in its velocity. Therefore, \(\Delta v = \frac{1.0 \times 10^{-27} \mathrm{kg \cdot m/s}}{1.0 \times 10^{-16} \mathrm{kg}} = 10^{-11} \mathrm{m/s}\). So, the change in the particle's velocity when it absorbs a photon of light with a wavelength of \(660 \mathrm{nm}\) is \(10^{-11} \mathrm{m/s}\).