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Problem 33 Medium Difficulty

The resolving power of a microscope is proportional to the wavelength used. A resolution of $1.0 \times 10^{-11} \mathrm{m}(0.010 \mathrm{nm})$ would be required in order to see an atom. (a) If electrons were used (electron microscope), what minimum kinetic energy would be required of the electrons? (b) If photons were used, what minimum photon energy would be needed to obtain $1.0 \times 10^{-11} \mathrm{m}$ resolution?

Answer

a. 15 \mathrm{keV}
b. 1.2 \times 10^{5} \mathrm{eV}

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Top Physics 103 Educators
Elyse G.

Cornell University

Marshall S.

University of Washington

Zachary M.

Hope College

Jared E.

University of Winnipeg

Video Transcript

Okay, So in this problem, we told the resolving power is proportional the wavelength. And so basically, um, we want to figure out ah, various questions related to that. So the 1st 1 is a kinetic energy required of electrons if they were toe have this required way playing. So it's 11 times 10 to the minus 11 meters, and we want to get the kinetic energy of the electrons. And so basically, we want to use Lambros. Each overpay p is equal to, um this crowd of two AM times a kinetic energy just because ah, catic energy is p squared over two hours. So just manipulating that so then we can solve for the Connecticut I g weaken say lam the square and divided by age equals one over for groups. So bring the age over square both, and you have to, um, Times K is that something you can get the, um, just continuing to algebraic lee manipulate. You got chaotic. Energy is equal to each squared over lambda squared, Um, times one over to Adam. And so go ahead and evaluate this for the mass of an electron. So let's do that. Each squared. I'm not squared one point. Oh, time seven minus 11. We're gonna square that and then were provided by two times 9.1 on. I have 10 to the minus, um, 31 kilogram could just a mass of the electron. Or that I got two point. So I got an answer in jewels, and I'm to convert it. Tiu be by dividing by 1.62 times kind of a minus 19. Um, And then with that, I got 15,000. Maybe I could just kind of used to a full number, I guess. Well, yeah, Well, I'll just ride 15,000. You ve great. And, ah, we wanted no foe times. What would the energy be? What would the energy needed be? So here is a this kind of little snake. Well, okay. And then now we want to get used the same lambda. So this lambda will come. I kind of come into this section as well. Um, we want to get the energy is each see over lambda for a proton. And so we evaluate that you're seeing divided by Lambda. Um then what? We can't iss Let's go ahead and go to my calculator you see, I really am the no converted todo a swell while it's really high, so I basically get 124 1000 bebe.

University of Washington
Top Physics 103 Educators
Elyse G.

Cornell University

Marshall S.

University of Washington

Zachary M.

Hope College

Jared E.

University of Winnipeg