This is a question from my textbook ---> \"An experimenter wishes to arrange a t

Question

This is a question from my textbook ---> "An experimenter wishes to arrange a two-slit ex- periment with 3-eV electrons so that the n = 1 maxi- mum occurs at 15°. What will his slit separation, d, have to be? "

Here is the solution:

THE QUESTION I NEED ANSWERED: How did they know 3 eV was Kinetic Energy and not total energy (E) in the problem? I used E^2 = (pc)^2 + (mc^2)^2 and plugged in the 3 eV into my E because it didnt state if the energy was kinetic or not. So how do u tell if its not stated in the problem?

KNOWN: Kinetic energy of electron, 3 ev Mass of electron, m 0.51 MeV /c Comment Step 2 of 5 A FIND: Find the separation between the two slits Comment Step 3 of 5 A SOLUTION The relativistic equation is The total energy is given by R R+m,c2 So, from relativistic equation

Explanation / Answer

Total enegy E is provided by the equation E^2 = (pc)^2 + (mc^2)^2

The first term (pc) is the kinetic energy and the second term (mc^2) is the energy equivalent of mass. Lets calculate the energy equivalent to the mass of the electron at rest

me = 9.11×10-31 kg; c=3 × 108 m/s. mec2 = 511.78 keV. This energy is many order of magnitude higher than the energy provided in the problem.

From this, we can understand that the energy provided (3eV) is only the kinetic energy and not the total energy of the electron. Also in this case of double slit diffraction, only the kinetic energy of the electron will take part. The energy equivalent of the mass of the electron won't play a role because this is not a nuclear reaction where the internal energy of the electron will take part.

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