Arrive at Heisenberg’s uncertainty principle with single slit electron diffraction. An electron has a speed of 300n/sec with uncertainty of 0.01 %. Find the accuracy in its position.

#### Solution

SINGLE SLIT ELECTRON DIFFRACTION : THE WAVE CHARACTRISTICS OF AN ELECTRON

⦁ Consider an electron moving in ‘x’ direction with a velocity ‘V_{x}' and an initial momentum of p_{x} = mv_{x} incident on a narrow slit of width ‘d’.

⦁ The electron is diffracted through an angle θ and strikes the screen the screen at point Q_{1} or point Q_{2 }on either side of the central point O.

⦁ On the way to the screen the electron gains a y component of momentum ‘p_{y}'. As a result it reaches the point Q_{1 }with a resultant momentum of p.

⦁ it is seen that p_{y } = psinθ, which varies with the angle of diffraction θ.

⦁ As there is no force acting in x-direction on the electron , p_{x } remains constant. Therefore the inaccuracy in the measurement of momentum arises from p_{y}.

The maximum uncertainty in the measurement of momentum can be the momentum itself.

Therefore, ∆p_{y} = p_{y } = psinθ. …………………………………(1)

⦁ For θ small it can be assumed that Q_{1 }

is the first minimum of the electron diffraction

pattern. In that case the condition for first minimum is:-

d sinθ = λ . …………………………………..(2)

⦁ From (1) and (2) it can be written as` ∆p_y = pλ/ d` ………………………………..(3)

⦁ On the other hand the electron needs to pass through any point of the slit , to be diffracted. Therefore the inaccuracy in determining the position of the electron in very small given by

`∆y_m = d` …………………………..(4)

⦁ From (3) and (4) it is found that

` ∆y_m ∆p_ yma = d.pλ/d = pλ`

`∆y_m.∆p _(yma) = h`

This verifies the uncertainty principle.

NUMERICAL:-

Given Data :- V = 300m/sec , ` ∆v /v= 0.01 %`

Formula :- ∆x.∆p ≥ ħ

Calculations :- ∆x.m.∆p ≥ ħ

`∆v = 300×(0.01)/(100) = 0.03`

`∆x ≥ ħ/ (m∆v) ≥ ( 6.63 ×10^( -34))/(2×3.14×9.1×0.03×10^(-31))`

≥ 3.8 ×10^{-3}

Therefore uncertainty in position = 3.8 ×10^{-3 }m