A beam of electrons is accelerated from rest through a potential difference of 0.200 kV and then passes through a thin slit. When viewed far from the slit, the diffracted beam shows its first diffraction minima at ± 13.6 ∘ from the original direction of the beam.

Do we need to use relativity formulas? Select the correct answer and explanation.

a. No. The electrons gain kinetic energy K as they are accelerated through a potential difference V, so Ve=K=mc2/(γ−1). The potential difference is 0.200 kV , soVe= 0.200 keV. Solving for γ and using the fact that the rest energy of an electron is0.511 MeV, we have γ–1=(0.511MeV)/(0.200keV) so γ−1>>1 which means that we do not have to use special relativity.
b. Yes. The electrons gain kinetic energy K as they are accelerated through a potential difference V, so Ve=K=(γ−1)mc2. The potential difference is 0.200 kV , soVe= 0.200 keV. Solving for γ and using the fact that the rest energy of an electron is0.511 MeV, we have γ–1=(0.200keV)/(0.511MeV) so γ<<1 which means that we have to use special relativity.
c. Yes. The electrons gain kinetic energy K as they are accelerated through a potential difference V, so Ve=K=mc2/(γ−1). The potential difference is 0.200 kV , soVe= 0.200 keV. Solving for γ and using the fact that the rest energy of an electron is0.511 MeV, we have γ–1=(0.511MeV)/(0.200keV) so γ>>1 which means that we have to use special relativity.
d. No. The electrons gain kinetic energy K as they are accelerated through a potential difference V, so Ve=K=(γ−1)mc2. The potential difference is 0.200 kV , soVe= 0.200 keV. Solving for γ and using the fact that the rest energy of an electron is 0.511 MeV, we have γ–1=(0.200keV)/(0.511MeV) so γ−1<<1 which means that we do not have to use special relativity.

Part B

How wide is the slit?