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

Answer:

v= 103.5 V; energy =1.65 x 10^-17

Explanation:

the deflected energy eV sin θ

Answer 2

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!