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 = 
E0E2 = E1 - ᐃE2 = E1 -
E1 = 
E0 and so on ⇒ ᐃE =
= 
E0 = 
E0E0 = 6.561 MeV, ᐃE = (6.561 - 0.001) MeV ⇒
= 1 - 
⇒ n = 4Create a free account to view solution
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