Electron Velocity Diffraction Grating Problem

AI Thread Summary
To solve the diffraction problem, the equation d*sin(theta) = m*lambda is used, where d is the slit distance, theta is the angle of the first maximum, and m is the order of the maximum. Given a slit distance of 0.01 mm and an angle of 3.6 degrees, the wavelength of the visible light is calculated to be approximately 6.279 x 10^-7 m. To achieve the same diffraction pattern with electrons, the De Broglie wavelength formula (lambda = h/mv) is applied. By rearranging this formula, the velocity of the electrons needed to match the diffraction pattern is found to be about 1.16 m/s. The theoretical approach is confirmed, though numerical accuracy should be verified.
Amad27
Messages
409
Reaction score
1

Homework Statement


Suppose that visible light incident on a diffraction grating with slit distance (space) of $0.01*10^{-3}$ has the first max at the angle of $3.6^{o}$ from the central peak. Suppose electrons can be diffracted with this same grating, which velocity of the electron would create the same diffraction pattern as this visible light?

Homework Equations



d*sin(theta) = m(lamba)

The Attempt at a Solution



I am thinking of using equations like $d\sin(\theta) = m\lambda$, but I am not sure where the angle comes into play here.

Obviously,

$(0.01 mm)(\sin(3.6)) = m\lambda$, but this doesn't help much?
 
Physics news on Phys.org
Amad27 said:
Suppose that visible light incident on a diffraction grating with slit distance (space) of $0.01*10^{-3}$ has the first max at the angle of $3.6^{o}$ from the central peak. Suppose electrons can be diffracted with this same grating, which velocity of the electron would create the same diffraction pattern as this visible light?

Homework Equations



d*sin(theta) = m(lamba)

The Attempt at a Solution



I am thinking of using equations like $d\sin(\theta) = m\lambda$, but I am not sure where the angle comes into play here.

Obviously,

$(0.01 mm)(\sin(3.6)) = m\lambda$, but this doesn't help much?

use the grating equation to find out the wavelength of the diffraction pattern -given out with light.
suppose you wish to have the same pattern with electron beam - then the wavelength associated with electron should be same.
can the the electron have wave property?
if it can then how wavelength depends on its velocity/energy/momentum?
the lambda should be related with velocity
 
Amad27 said:
equations like $d\sin(\theta) = m\lambda$,

A friendly tip: to show your equations inline with text, enclose your LaTeX equations in ##, not $. To show them as separate "paragraphs", use $$.
 
Thanks.

Okay:$$0.01\sin(3.6) = m\lambda$$ so m=1, and thus

$$\lambda = 6.279*10^{-7} m$$

Then by De Broglie,

$$\lambda = h/mv$$ Thus,

$$v = \frac{6.63*10^{-34}}{(9.1*10^{-31})(6.279*10^{-7})} = 1.16 m/s$$

Is this right?
 
Amad27 said:
Thus,

v=6.63∗10−34(9.1∗10−31)(6.279∗10−7)=1.16m/s​
v = \frac{6.63*10^{-34}}{(9.1*10^{-31})(6.279*10^{-7})} = 1.16 m/s

Is this right?
i have not checked your numbers but theoretically the idea is same
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top