CP A beam of 40-eV electrons travelling in the +x direction passes through a slit that is parallel to the y-axis and

5.0 mm wide. The diffraction pattern is recorded on a screen 2.5 m from the slit. (a) What is the de Broglie wavelength of the

electrons? (b) How much time does it take the electrons to travel from the slit to the screen? (c) Use the width of the central diffraction pattern to calculate the uncertainty in the y-component of momentum of an electron just after it has passed through the

slit. (d) Use the result of part (c) and the Heisenberg uncertainty principle (Eq. 39.29 for y) to estimate the minimum uncertainty

in the y-coordinate of an electron just after it has passed through the slit. Compare your result to the width of the slit.

de Broglie wavelength is an important concept while studying quantum mechanics. The wavelength (λ) that is associated with an object in relation to its momentum and mass is known as de Broglie wavelength. A particle's de Broglie wavelength is usually inversely proportional to its force

Knowns

From equation 39.1, the de Broglie wavelength of :1 particle with momentum

p is given by:

\\"linear b (3.626 2‘: I(l 3* It is Planck 0011810111,.

From equation 38.12. the angle of the ?rst minimum (m = 1), When a beam

of wavelength ,\ is (lifractcd through a slit of width (,1 is given by:

]_\

Givens

The kinetic energy of the electrons is K = 40.08V , the width of the slit is a ,_\ and the distance between the

slit and the screen is R = 2.50m -

(a) First, we convert the kinetic energy of the electron from electron volts to joules as follOWS:

We know that the non relativistic kinetic energy is given by:

And the non relativistic momentum is given by:

Combining the two equations, we get:

Now, we plug our values for K and me, so We get the momentum of the electron:

Now, we plug this value into equation (1), so we get the de Broglie Wavelength of the electrons:

Now, we plug this value into equation (1), so we get the de Broglie Wavelength of the electrons:

(b) First, we calculate the speed of the electrons, so We solve equation (3) for v and plug our values for me and p, so We get

the speed of the electrons:

This speed is very small compared to the speed of light, so using the nonrelativistic form is acceptable.

Therefore Using kinematics, the time t it takes for the electrons to travel the distance R is:

(c) The width of the central diffraction pattern is twice the distance from the central maximum and the ?rst minimum, 10 : 2y-

From Fig- 38.17, the distance betWeen the central maximum and the ?rst minimum is given by:

Where 0 is too small-

Substituting for 6 from equation (2), We get:

Substituting for our given values, we get:

I've beam is directed along the m-axis, but many electrons spread near the center; So, the electrons have a y-component of

relocity vy-

I've electrons spread along the y-axis over a total distance of U) with an uncertainty of y—component velocity Avy-

Ne can get this uncertainty in the y-component of velocity as follows

Where t is the time it takes for the beam to travel from the slit to the screen which has been calculated in part (b)

Where t is the time it takes for the beam to travel from the slit to the screen which has been calculated in part (b)-

80, the uncertainty in the y-component of momentum of the electrons is:

mew

Now, we plug our values for me, 11) and t, so we get:

From equation 39.29. the Heisenberg uncertainty principle for y-component

of momentum and y-coordinate is:

So, the minimum uncertainty in the y-coordinate of the electrons is =0.20 mu.