Question

An electron of energy 200 ev is passed through a circular hole of radius what is uncertainty introduced in the angle of emergence

Answer 1

in think you question is

question :An electron of energy 220 eV is passed through a circular hole of radius 2 x 10-4 cm. What is the uncertainty introduced in the angle of emergence?

Answer:

Explanation:

The energy E of the electron is related to its momentum

E= $\frac{{p}^2 }{2m}$ where p is the momentum and m(e) is mass of the electron

therefore momentum

p = $\Sqrt{(2.m(e)}.{E)}$

The Heisenberg Uncertainty Principle states that

Delta(y). delta(p) = of the order of h the planks constant

if the radius of the slit is taken of the order of $2. {10}^{-4} cm = 2. 10^{-6} m$

the maximum uncertainty is of the order of width which is (2.radius).

D =$4. 10^ {-6} m$

as the slit is circular the electron de Broglie wave will diffract and

and uncertainty in angle will be of the order of 1.22( lambda/D) where D is the diameter of the hole.

the wavelength lambda = h/momentum

Delta(theta) = 1.22. (h/p). (1/D)

Delta(theta) =$\frac{ 1.22.  h}{ (2.m(e).E))}}.\frac{1}{D}$)

putting the values

= $\frac{1.22. ( 6.62 x 10-34 J s. ) } { {\Sqrt(2. x 9.1 .10^{-27} kg x 220 x 1.6 x 10^{-19} J)} .(4. 10^ {-6 m})}$

angular uncertainty = of the order of 1.4 x 10^-5 degree

Answer 2

Sol is in the attachment