Just need part b A beam of electrons is accelerated from rest through a potentia
ID: 1639159 • Letter: J
Question
Just need part b
A beam of electrons is accelerated from rest through a potential difference of 0.300 kV and then passes through a thin slit. When viewed far from the slit, the diffracted beam shows its first diffraction minima at plusminus 13.1 degree from the original direction of the beam. difference is 0.300 kV, so Ve = 0.300 keV. Solving for gamma and using the fact that the rest energy of an electron is 0.511 MeV, we have gamma-1 = (0.511 MeV)/(0.300 keV) so gamma > > 1 which means that we have to use special relativity. No. The electrons gain kinetic energy K as they are accelerated through a potential difference V, so Ve = K = mc^2/(gamma-1). The potential difference is 0.300 kV, so Ve = 0.300 keV. Solving for gamma and using the fact that the rest energy of an electron is 0.511 MeV, we have gamma-1 = (0.511 MeV)/(0.300 keV) so gamma - 1 > > 1 which means that we do not have to use special relativity. Yes. The electrons gain kinetic energy K as they are accelerated through a potential difference V, so Ve = K = (gamma - 1)mc^2. The potential difference is 0.300 kV so Ve = 0.300 keV. Solving for gamma and using the fact that the rest energy of an electron is 0.511 MeV, we have gamma-1 = (0.300 keV)/(0.511 MeV) so gammaExplanation / Answer
B)
We are given,
V =0.300kV = 0.300*10^3 V
Let us first calculate the velocity and wavelength of electron in the beam
Using equation,
KE = 1/2mv^2
q*V = 1/2mv^20
1.6*10^-19*0.300*10^3 = ½*9.1*10^-31*v^2
v= 1.027*10^7 m/s
By de-Broglie hypothesis,
=h/p
=h/mv
=(6.62*10^-34)/(9.1*10^-31*1.027*10^7)
= 7.083*10^-11 m
Now we use diffraction formula for single slit,
dsin=m
d= m/sin
=13.1 deg, for 1st minima m=1
Plugging values,
d=(1*7.083*10^-11)/(sin13.1)
slit width = d= 3.13*10^-10 = 3.13 A
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.