Calculate the de-broglie wavelength of an alpha particle accelerating through a potential difference of 200 volts given:- mass of alpha particle = $6.68×10^{(-27)}kg. $

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 = 0.09 A°.$

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

$λ_E = 1.65 A°.$

On the other hand since the electron energy is

$ E = 1/2 mv^2 = eV$

The wavelength associated with an electron wave can be written as

$λ_T= h/(√2mE)= h/(√2meV)= 12.25/(√V) $

here the accelerating potential V = 54 volts for which the theoretical electron wavelength is found to be

$λ_T = 1.67 A°.$

Which is very close to the experimental value , $ λ_E.$

This agreement confirms de Broglie hypothesis of matter waves.

NUMERICAL:

Given data : - $m_p = 6.68 × 10^{(-27)}kg, q_α=4e=6.4 ×10^{(-19)} C$

$ V = 200 volts.$

Formula : - $1/2 mv^2=q_α V , \hspace{1cm} λ= h/mv$

Solutions :- - $ 1/2 mv^2=q_α V , \hspace{1cm} v= √(2q_α V)/m$

$ λ= h/(√2q_α mV)$

$ = (6.63 × 10^{(-34)})/(√2×6.4×10^{(-19)}×6.68×10^{(-27)}×200)$

$ =0.507$

Answer : de Broglie wavelength of an alpha particle

$ λ_α = 0.507 A°.$