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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