Achieve 15

[April_Tello_2H]

You use an electron microscope in which the matter wave associated with the electron beam has a wavelength of 0.0231 nm. What is the kinetic energy of an electron in the beam, expressed in electron volts?

I really need help with this question. I know I would have to cover the nm to then have it be (2.31 x 10^-11). I would then use the DeBroglie equation but solve for v in which it would be

v= m*v / Lambda

when plugging in my number I have

(6.626 * 10^-34) / (9.11 * 10 ^-31) (1.29 * 10^-11)

Yes I understand that solving this equation I would get the velocity then used into finding the kinetic energy BUT I cannot get my calculation for V to have 10^-16 and was wondering If someone can input this in the calculator and find my V ? I just need this part and cannot get my calculation for V.

Please and Thank you

[Christian_Lee_2K]

Hi!

I'll walk you through what I had done. I used the equation v = h/(m(lamdha)) and substituted it into the kinetic energy equation (1/2)mv^2, giving you KE = (1/2) * m * (h/(m(lamdha)))^2. Now, the values of each should be as follows:

m = mass of an electron = 9.109 x 10^-31 kg
h = planck's constant = 6.626 x 10^-34 J
lamdha = 0.0369 x 10^-9 m

My answer came to be 1.77 x 10^-16 J. Then, I convert this to electron volts with 1 eV = 1.602 *10^19 J so the answer will be 1.10 x 10^3 eV.

In response to your work, I believe that you're missing a set of paranthesis.
(6.626 * 10^-34) / ((9.11 * 10 ^-31) (1.29 * 10^-11))
In your prior input, the calculator had divided h by m of electron then multiplied it by the (1.29 * 10^-11)