CP A beam of electrons is accelerated from rest and then passes through a pair of identical thin slits that are 1.25 nm

apart. You observe that the first double-slit interference dark fringe occurs at 18.0 from the original direction of the beam

when viewed on a distant screen. (a) Are these electrons relativistic? How do you know? (b) Through what potential difference were the electrons accelerated?

From equation 36.13, when a beam of Wavelength A is diffracted from double slits separated by distance d, the condition of a

bright fringe is:

Where am is the angle of line from centre of the distance betWeen the two slits to mth bright fringe on screen-

From equation 39.], the de Broglie wavelength of a particle of momentum p is given by:

Step 2 2 of 5

Givens

The distance between slits is and the angle of the ?rst dark fringe is -

Calculations

(a) Equation (1) show$ the condition of maxima, and we know that there is a minimum betWeen every tWO maxima;

So, we can express the condition of dark fringe as follows

For the ?rst dark fringe, m = 1-

So, we plug our values for 61 and (1 into equation (3), so we get the waveelngth of the electrons:

Then, we plug this value into equation (2), so we get the momentum of the electrons:

The nonrelativistic momentum is given by:

So, we substiutute for me and p, so We get the speed of the electrons:

This speed is very small compared to the speed of light 6 = 3.00 X 108m/s , so Therefore these electrons are nonrelativistic-

(b) The nonrelativistic kinetic energy is given by:

So, We plug our values for me and 1), so we get the kinetic energy of the electrons:

When an electron is accelerated through a potential difference V, it gains kinetic energy of 6V;

When an electron is accelerated through a potential difference V, it gains akinetic energy of 8V;

Thus, the potential difference through which the electrons are accelerated is:

e 1.60 x 10-19 C

Therefore the potential difference is V = 2.52 v