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Answers
here, wavelength, λ = 10^-15 m
from De-broglie's wavelength, λ = h/p
where h is Plank's constant.i.e., h = 6.63 × 10^-34 J.s
and p is momentum of electron associated with wavelength λ
so, p = h/λ
= (6.63 × 10^-34)/(10^-15)
= 6.63 × 10^-19 Kgm/s
rest mass energy of electron, ∆mc² = 0.511 MeV = 0.511 × 10^6 × 1.6 × 10^-19J
= 0.511 × 1.6 × 10^-13 J
now use the relativistic formula for the energy of electron.i.e., E =
= √{(3 × 10^8)² × (6.63 × 10^-19)² + (0.511 × 1.6 × 10^-13)²}
= 1.989 × 10^-10J
we know, 1 eV = 1.6 × 10^-19J
so, E = 1.989 × 10^-10/(1.6 × 10^-19)
= 1.24 BeV
it is clear that energy acquired by electron from the given accelerator must have been of the order of a few BeV.
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