Achieve Question 13

[205992763]

Hi! I got the first part of the question and know that the energy of the photon is 1.03 x 10^-19 but I cannot find the energy of the electron. If someone could help me out that would be greatly appreciated. Thank you so much!

The E.coli bacteria is about 1.9 micrometers long. Suppose you want to study it using photons of that wavelength or electrons having that de Broglie wavelength.

[Kaya Lu 2K]

Hi! How I solved part b was that I first used the de Brogile equation to solve for the momentum (p) of the electron. p = h/wavelength

From there, I used the p I obtained to solve for velocity (v = p/m with m = mass of an electron).

Finally, I plugged the mass of an electron as well as the v that I obtained into the formula for the kinetic energy of an electron, E[electron] = (1/2)mv^2.

I hope this helps!

[noree_jauregui]

To find the energy of an electron, you use the equation wavelength is equal to h/p. You use p=mv to plug into the first equation. You then move the equation around to solve for v instead of wavelength since you need v to solve for the energy. After you find the v, you can plug it into the equation E(kinetic)=1/2 (mv^2) to get the energy of the electron.

[Kait Nguyen]

To find the energy of the electron, you need to use KE=1/2mv^2

So to start, use de broglie’s (λ =h/mv) to find the velocity plugging in what you know
1.9x10^-6 (this is length your the bacteria in m)=6.626x10^-34/(9.11x10^-31)(v)

Then plug in to find the KE, which will be the energy of the electron

Hope this helps!