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.

Answer 1

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?

Answer:

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:

$\frac{1}{2}mv^2 = eV$

Making v the subject of the formula; we have:

$v = \sqrt{\frac{2eV}{m}}$

where our constants are

m = $9*10^{-31} kg$

e = $1.60*10^{-19} C$

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

substituting our parameters; we have:

$v = \sqrt{\frac{2(1.60*10^{-19}C(400V)}{9.1*10^{-31}kg}}$

$v= 11.90 *10^6ms^{-1$

From De broglie wave equation;

$\lambda = \frac{h}{mv}$    --------------Equation(1)

Davisson and Germer Experiment also shows that;

$asin \theta=n \lambda$

making $\lambda$ the subject of the formula; we have:

$\lambda = \frac{asin \theta}{n}$  --------------- Equation (2)

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

$\frac{asin \theta }{n} =\frac{h}{mv}$

${asin \theta } =n\frac{h}{mv}$

$a=\frac{n}{sin \theta}(\frac{h}{mv} )$       -----------  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 = $6.63 *10^{-34} kgm^2s^{-1}$

n = 1

m = $9.1*10^{-31}kg$

$v= 11.90 *10^6ms^{-1$

θ  = 15. 1°

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

$a=\frac{6.63*10^{-34}*(1)}{(9.1*10^{-31})(11.9*10^6)(sin15.1)}$

$a=2.35*10^{-10}m$

Converting m to nm; we have

$a=(2.350*10^{-10}m)(\frac{1nm}{10^{-9}m} )$

$a = 2.35*10^{-1}nm$

a = 0.235 nm

∴ the width of the slit = 0.235 nm

I hope that helps alot!

Answer 2

The width of the slit in the given question scenario is; a = 0.2357 nm

We are given;

Potential difference; V = 0.4 Kv = 400 V

angle of diffraction; θ = 15.1°

Now, from law of conservation of energy, we can say that;

¹/₂mv² = eV

where;

m is mass of electron

v is speed of electron

e is charge of electron

V is Potential difference/Voltage

Now, let us make the velocity the subject of the equation;

v = √(2eV/m)

where;

m = 9.1 × 10⁻³¹ kg

e = 1.6 × 10⁻¹⁹ C

Thus;

v = √(2 × 1.6 × 10⁻¹⁹ × 400/(9.1 × 10⁻³¹)

v = 11.86 × 10⁶ m/s

From a combination of the formula of wavelength in De broglie's wave equation and that in the Davisson - Germer experiment on the wave nature of electrons. we have;

a = nh/(mv sin θ)

where;

a is width of slit

n is  order of diffraction = 1

h is Planck's constant = 6.626 × 10⁻³⁴ J.s

m is mass of electron = 9.1 × 10⁻³¹ kg

v is speed of electron = 11.86 × 10⁶ m/s

θ is angle of diffraction

The question wants us to find width of slit and so;

a = (1 × 6.626 × 10⁻³⁴)/(9.1 × 10⁻³¹ × 11.86 × 10⁶ × sin 15.1)

a = 0.2357 × 10⁻⁹ m

a = 0.2357 nm

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