Double-Slit Interference (Thank you in Advance)

[kjsingh1]

Homework Statement

A beam of electrons is accelerated from rest and then passes through a pair of identical thin slits that are 2 nm apart. You observe the first double-slit interference dark fringes occurring at ±20◦ from the original direction of the beam when viewed on a distant screen.

(A) Calculate the speed of the electrons.

(B) Through what potential difference were these electrons accelerated?

Homework Equations

sin(θ)=((m+1/2)λ)/a
(a=width)

f=c/λ

v=fλ

K=.5mv^2

V(Potential Difference) = K/e

The Attempt at a Solution

(A)

sin(θ)=((m+1/2)λ)/a
λ=asin(θ)/(m+1/2)

λ=(2*10^-9m)sin(20)/(1+1/2)
λ=4.56*10^-10

V=c=speed of light? since f=c/λ

(B)

K=.5mv^2=.5(9.1*10^-31)(3*10^8)^2
K=4.095*10^-14

V=K/e
V=(4.095*10^-14)/(1.602*10^-19)

V=255617.98 ?

 

[ehild]

The Attempt at a Solution

(A)

sin(θ)=((m+1/2)λ)/a
λ=asin(θ)/(m+1/2)

λ=(2*10^-9m)sin(20)/(1+1/2)
λ=4.56*10^-10

V=c=speed of light? since f=c/λ

m=0 at the first minimum: The difference between the distances traveled by the two waves is just half wavelength.

Electron are particles, but at the same time they have wave properties: How is the wavelength of the electron defined?
The electron has rest mass, so it never travels with the speed of light.

ehild

 

[kjsingh1]

Oh yeah, totally forgot.

So, wavelength = h/mv

Meaning Velocity=h/mλ

So, the correct answers would be:

λ=((2*10^-9)sin(20)) / (1/2) = 1.368*10^-9 m

Vel.=(6.626*10^-34) / ((9.1*10^-31)(1.368*10^-9)) = 5.32*10^5 m/s

Then,

K=(1/2)(9.1*10^-31)(5.32*10^5 m/s)^2 = 1.289*10^-19

V(potential diff.) = (1.289*10^-19) / (1.602*10^-19) = .805 V ??

Thank you so much, I am really starting to understand this concept.

 

[kjsingh1]

Can anyone Please help me further with this??

I really need this!

 

[ehild]

What is your problem? ehild