## 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

Question

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?

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5 months 2021-09-03T18:07:41+00:00 1 Answers 6 views 0

## Answers ( )

a) the correct answer is d
, b)    a = 3.69 10⁻⁸ m

Explanation:

a) to see which answer is correct let’s sketch the solution to the problem

ΔU = K

K = (γ -1) mc²

(γ -1) = K / mc²

(γ-1) = ΔU / mc²

(γ-1) = e V / mc²

If we work in electron volt units ΔU = V    [eV]

(γ-1) = V / mc²

(γ -1) = 0.2 10³ / 0.511 10⁶

(γ -1) = 3.9 10⁻⁴

As it is very small, relativistic corrections are not necessary.

Checking the correct answer is d

b) let’s use De Broglie’s relationship to find the wavelength of electrons

λ = h / p = h / mv

Let’s look for the speed of electrons, for this we use the concept of energy conservation

Initial

Em₀ = ΔU = e ΔV

Final

Emf = K = ½ m v²

Em₀ = Emf

e ΔV = ½ m v²

v = √ (2 e ΔV / m)

v = √ (2 1.60 10⁻¹⁹ 0.2 10³ / 9.1 10⁻³¹)

v = √ (70.33 10¹²)

v = 8.39 10⁶ m / s

Much less than the speed of light

We replace

λ = 6.63 10⁻³⁴ / (9.1 10⁻³¹ 8.39 10⁶)

λ = 8.68 10⁻⁹ m

The diffraction is explained by the expression

a sin θ = m λ

The minimum occurs for m = 1

a = λ / sin θ

a = 8.68 10⁻⁹ / sin 13.6

a = 3.69 10⁻⁸ m