HW problem #25

$\lambda=\frac{h}{p}$

Jessica Manriquez 1H
Posts: 50
Joined: Wed Sep 30, 2020 9:47 pm

HW problem #25

Hi everyone! im kind of lost on how to start this problem and was looking for any guidance; thanks in advance. :)

The E.coli bacterium is about 2.0 μmlong. Suppose you want to study it using photons of that wavelength or electrons having that de Broglie wavelength.

What is the energy Ephoton of the photon? What is the energy Eelectron of the electron?

Posts: 97
Joined: Wed Sep 30, 2020 9:42 pm

Re: HW problem #25

The problem gives you the wavelength of the photon (2.5 μm) so you would use the E = hv equation but substitute v for c/wavelength to make the equation in terms of the wavelength to find Ephoton.

As for the energy of the electron, it asks for the energy of the electrons having a de Broglie wavelength, so you would use the de Broglie expression. Since velocity is the only unknown variable, you would solve for the velocity, and since the only energy equation that takes velocity into account is the kinetic energy equation (1/2mv^2), you would solve for the energy of the electron using that equation.

Hope this helps!

MinjooPark_3I
Posts: 105
Joined: Wed Sep 30, 2020 9:33 pm

Re: HW problem #25

Hi. I just did this problem!
So what I did was first use the equation E=hc/lambda in order to find the energy of the photon. Since we have the wavelength in mu meters, we would convert to meters and plug the values into the equation. This gives you the answer to the first question.

For the second question, we used de Broglie's equation lambda=h/mv but switched around the equation to solve for the velocity. I plugged in the wavelength in meters into the switched equation v=h/(m)(lambda). After getting my velocity, I plugged it into the equation Ek= 1/2mv^2 to get the energy of the electron.

I hope this helped!

Jessica Manriquez 1H
Posts: 50
Joined: Wed Sep 30, 2020 9:47 pm

Re: HW problem #25

Perfect!! thank you!

rhettfarmer-3H
Posts: 96
Joined: Wed Sep 30, 2020 9:59 pm

Re: HW problem #25

Hey this one is weird but dont worry i got you. So since you are given wave length as stated in the problem the first step is to find the energy per photon which is e=hc/lamanda. Then for the electron the relationship we have for velocity of Electron is debroglies which we take the h/lamanda =mv. All is given but V. Finally, to find the energy we use sweet old kinetic E which is 1/2(mv^2). Done deal.