Nuclear Physics and RadioactivityHard

Question

How many head - on elastic collisions must a neutron have with deuterium nuclei to reduce it energy from 6.561 MeV to 1 keV?

Options

A.4
B.8
C.3
D.5

Solution


After 1st collision ᐃE1 = E0, After 2nd collision ᐃE2 = E1, After nth collision ᐃEn = En -1.
Adding all the losess
ᐃE = ᐃE1 + ᐃE2 + ....... + ᐃEn = (E0 + E1 + ......+ En - 1) ; here E1 = E0 - ᐃE1 = E0 - E0 = E0
E2 = E1 - ᐃE2 = E1 - E1 = E0 and so on
⇒ ᐃE = = E0 = E0
E0 = 6.561 MeV, ᐃE = (6.561 - 0.001) MeV ⇒ = 1 - ⇒ n = 4

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