Hi Please help me with this problem! My final is tomorrow and I\'m so screwed. E

Question

Hi Please help me with this problem!
My final is tomorrow and I'm so screwed.

Electrons were started at rest and accelerated through an electricpotential so that they reached
a final velocity. They were then passed through a horizontal 5 nmslit. The first minimum in the number
of electrons detected on the other side of the slit is at an angleof 15º above the normal to the slit.
A) What was the potential through which the electrons werepassed?

B) what was the electrons velocity.


THANK YOU SO MUCH

Explanation / Answer

the rest energy of the electrons is Eo= 0.511 MeV = 0.511 * 10^6 eV = 0.511 * 10^6 *(1/1.6 * 10^-19) J A)we know that Eo= e * Vo e is the charge on the electron and Vo is thepotential difference through which the electron isaccelerated or Vo= (Eo/e) = [0.511 * 10^6 * (1/1.6 *10^-19)/1.6 * 10^-19] = (0.511 * 10^6/(1.6 * 10^-19)^2) B)the kinetic energy of the electron is Eo= (1/2)m * v^2 or v = (2Eo/m)^(1/2) m is mass of electron and has a value 9.1 * 10^-31 kg e is the charge on the electron and Vo is thepotential difference through which the electron isaccelerated or Vo= (Eo/e) = [0.511 * 10^6 * (1/1.6 *10^-19)/1.6 * 10^-19] = (0.511 * 10^6/(1.6 * 10^-19)^2) B)the kinetic energy of the electron is Eo= (1/2)m * v^2 or v = (2Eo/m)^(1/2) m is mass of electron and has a value 9.1 * 10^-31 kg

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