Question

The phenomenon of Brownian motion is the random motion of microscopically small particles as they are buffeted by the still smaller molecules of a fluid in which they are suspended. For a particle of mass $1.0 \times 10^{-16} \mathrm{kg},\( the fluctuations in velocity are of the order of \)0.010 \mathrm{m} / \mathrm{s} .$ For comparison, how large is the change in this particle's velocity when the particle absorbs a photon of light with a wavelength of \(660 \mathrm{nm}\) such as might be used in observing its motion under a microscope?

Short answer

Answer: The change in the particle's velocity is 10^{-11} m/s.

Step-by-step solution

Find the energy of the photon

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, \(E = \frac{(6.63 \times 10^{-34} \mathrm{Js})(3.0 \times 10^8 \mathrm{m/s})}{(660 \times 10^{-9} \mathrm{m})} = 3.01 \times 10^{-19} \mathrm{J}\).

Use conservation of momentum to find the change in velocity

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 \(10^{-11} \mathrm{m/s}\).