An electron has a speed of 300n/sec with uncertainty of 0.01 %. Find the accuracy in its position.

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 $∆p_y=p λ/d$ ………………………………..(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

$∆y_m ∆p_yma=d.p λ/d=pλ$

$∆y_m.∆p_yma=h$

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

$∆x ≥ħ/m∆v ≥ \frac{(6.63 ×10^{-34})}{(2×3.14×9.1×0.03×10^{-31} )}$

≥ $ 3.8 ×10^{-3}$

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