Week 2, 3, 4 Question 13

[205957986]

i cannot figure out the first part of this problem:

it states that the e coli bacterium is 2.4 um long, and is asking us to determine the energy (e photon) of the photon, but i cannot seem to calculate it right.

[Shaniya Kerns 4D]

In order to determine the energy of the photon would use the equation the equation for speed of light and rearrange it to be f=c/lambda, f being the frequency of waves, c being the speed of light, and lambda being wavelength. You would combine that equation with plancks equation E=hf by plugging in f to get, E=(hc)/(lambda), this equation is the equation you plug in your answers with in order to get the energy of photon. Therefore 2.4pm represents lambda or wavelength, h or planck's constant= 6.62608 × 10^–34 J.s, c or speed of light=2.99792 × 108 m.s−1. Hope this helps!