Q. 313.8( 5 Votes )

# Crystal diffracti

Answer :

Given:

Wavelength of probe, λ = 1 Å = 10^{-10}m

Mass of electron, m_{e} = 9.11 × 10^{-31}Kg

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.

Rate this question :

If light ofPhysics - Board Papers

A. How does one ePhysics - Board Papers

Explain with the Physics - Board Papers