Q. 313.8( 5 Votes )

# Crystal diffracti

Answer :

Given:

Wavelength of probe, λ = 1 Å = 10-10m

Mass of electron, me = 9.11 × 10-31Kg

The kinetic energy of an electron is given by: We can write,

mv = (2 × m × KE)0.5

P = (2 × m × KE)0.5 ...(1)

Where,

m = mass of electron

v = velocity of electron

p = momentum of particle

De-broglie wavelength is given by, Where,

λ = Wavelength

h = Planck’s constant

p = momentum

From equation (1) we can write, …(2)

From equation (2) we can write that,

KE = Putting the values in above equation we get,

KE = KE = 2.39 × 10-17 J

KE for an electron is 2.39 × 10-17 J.

Now for photon,

E’ = …(4)

Where,

E = energy of photon

h = Planck’s constant

c = speed of light

λ = wavelength of the photon

By putting values in equation (4 ) we get, E’ = 19.6 × 10-16 J

The photon will have more energy as compared to the accelerated electron for the same wavelength.

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