Answered: Bragg condition Distance between for…

Related questions

Question

In the 1920s Clinton Davisson and Lester Germer accidentally observed diffraction when electrons with 54 eV of energy were scattered off crystalline nickel. The diffraction peak occurred when the angle between the incident beam and the scattered beam was 50°. (a) What is the corresponding angle u relevant for Eq.? (b) The planes in crystalline nickel are separated by 0.091 nm, as determined by x-ray scattering experiments. According to the Bragg condition, what wavelength do the electrons in these experiments have? (c) Given the mass of an electron as 9.11 x 10^-31 kg, what is the corresponding classical speed vcl of the diffracted electrons? (d) Assuming the electrons correspond to a wave with speed vcl and wavelength λ, what is the frequency f of the diffracted waves? (e) Quantum mechanics postulates that the energy E and the frequency f of a particle are related by E = hf, where h is known as Planck’s constant. Estimate h from these observations. (f) Our analysis has a small flaw: The relevant wave velocity, known as a quantum phase velocity, is half the classical particle velocity, for reasons explained by deeper aspects of quantum physics. Re-estimate the value of h using this modification. (g) The established value of Planck’s constant is 6.6 x 10^-34 J . s. Does this agree with your estimate?

Bragg condition
Distance between
for constructive
adjacent rows in array
Wavelength
interference
2d sin 0 = mX
(т %3D 1, 2, 3, ...)
from an array:
Angle of line from surface of array to mth bright region on screen

Expert Solution

Step 1

Braggs condition for diffraction from crystalline planes is given as

$2 d \sin θ = n λ d i s t h e s p a c i n g b e t w e e n c r y s t a l p l a n e s θ i s t h e g l a n c i n g a n g l e, w h i c h i s m e a s u r e d f r o m t h e c r y s t a l p l a n e t o t h e i n c i d e n t b e a m n i s t h e o r d e r o f d i f f r a c t i o n λ i s t h e w a v e l e n g t h o f r a d i a t i o n$

As given, diffraction is produced when the angle between the incident and scattered beams is 50

$T h e g l a n c i n g a n g l e θ i s e q u a l t o h a l f t h e a n g l e b e t w e e n t h e i n c i d e n t a n d s c a t t e r e d b e a m s S o, θ = \frac{50}{2} θ = 25 °$

Step 2

$F o r d = 0.091 n m = 0.091 \times 10^{- 9} m C o n s i d e r i n g a f i r s t o r d e r d i f f r a c t i o n, n = 1 n λ = 2 d \sin θ λ = 2 \times 0.091 \times 10^{- 9} \times \sin 25 λ = 7.7 \times 10^{- 11} m$

This is the required wavelength of the incident beam

Step by step

Solved in 4 steps

Bragg condition Distance between for constructive adjacent rows in array Wavelength interference 2d sin 0 = mX (т %3D 1, 2, 3, ...) from an array: Angle of line from surface of array to mth bright region on screen