why in Davisson and german experiment, he used nickel?? is nickel only suitable for this?? please explain
Answers
According to de Broglie’s hypothesis, a beam of material particle must possess wave like characteristics and should undergo phenomena like reflection , refraction, interference, diffraction and polarization as the ordinary light waves do. The first experimental verification of wave nature of atomic particles was provided by davisson and germer.
DAVISSON-GERMER EXPERIMENT.
In this experiment , from a hot filament F electrons are accelerated by a small voltage V to strike a target made of a single nickel crystal.
The electrons are scattered off the crystal in all directions.
Help of an scattered electrons in all directions can be recorded with the help of an electron detector which can be rotated on a circular graduated scale.
For any accelerating voltage ,V the scattering curve shows a peak or maximum in a particular direction.
It is found that for an accelerating voltage of 54 volts a very large number of electrons are scattered at a particular angle , ∅=50°.
It is assumed that the electron undergoes diffraction and the peak represents the first order spectrum at an angle of 50°.
The diffraction effects is explained as follows.
The atomic planes of the nickel crystal act like the ruling of a diffraction grating.
The interatomic distances of a nickel crystal is known to be a = 2.15 A°.
The interatomic spacing of a nickel crystal is
d=0.09A°.
1
Since, it is assumed that the electrons undergo diffraction they must follow braggs law,
2dsinθ = nλ
It is seen that the glancing angle θ = 65°. Assuming the order of diffraction , n= 1 the electron wavelength is experimentally calculated as
λE=1.65A°.
On the other hand since the electron energy is
E=1/2mv2=eV
The wavelength associated with an electron wave can be written as
λT=h/(√2mE)=h/(√2meV)=12.25/(√V)
here the accelerating potential V = 54 volts for which the theoretical electron wavelength is found to be
λT=1.67A°.
Which is very close to the experimental value , λE.
This agreement confirms de Broglie hypothesis of matter waves.
2
NUMERICAL:
Given data : - mp=6.68×10(−27)kg,qα=4e=6.4×10(−19)C
V=200volts.
Formula : - 1/2mv2=qαV,λ=h/mv
Solutions :- - 1/2mv2=qαV,v=√(2qαV)/m
λ=h/(√2qαmV)
=(6.63×10(−34))/(√2×6.4×10(−19)×6.68×10(−27)×200)
=0.507
Answer : de Broglie wavelength of an alpha particle
λα=0.507A°.
to speed up the reaction that takes place