A beam of electrons is accelerated from rest through a potential difference of 0.400 kV and then passes through a thin slit. When viewed far from the slit, the diffracted beam shows its first diffraction minima at ± 15.1 ∘ from the original direction of the beam.

Here is the full question:

A beam of electrons is accelerated from rest through a potential difference of 0.400 kV and then passes through a thin slit. When viewed far from the slit, the diffracted beam shows its first diffraction minima at ± 15.1° from the original direction of the beam.

How wide is the slit?

0.235 nm

Explanation:

Given that:

The potential difference (V) = 0.400 kV = 400 V

angle  of diffraction (θ) = 15.1°

Using the conservation energy to determine the seed of the electron; we have:

Making v the subject of the formula; we have:

where our constants are

m =

e =

given potential difference (V) = 0.400 kV = 400 V

substituting our parameters; we have:

From De broglie wave equation;

   --------------Equation(1)

Davisson and Germer Experiment also shows that;

making the subject of the formula; we have:

 --------------- Equation (2)

Equating equation (1) and (2); we have:

      -----------  Equation(3)

where;

a = width of the slit

n = order of diffraction

θ = angle of diffraction

Since we were told that when the beam of electrons were viewed from the slit, the diffracted beam shows its first diffraction minima ;

then (n) = first order = 1

where: h =

n = 1

m =

θ  = 15. 1°

Substituting our values into equation (3); we have:

Converting m to nm; we have

a = 0.235 nm

∴ the width of the slit = 0.235 nm

I hope that helps alot!

answer: point c: b. 1/2 potential, 1/2 kinetic, point a: 3/4 of its potential energy, point d or b: 3/4 potential, 1/4 kinetic

explanation: i cant explain. it's difficult to explain.