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On an average, a neutron loses half of its energy per collision with a quesi-free proton. To reduce a `2 MeV` neutron to a thermal neutron having energy `0.04 eV`, the number of collisions requaired is nearly .A. `50`B. `52`C. `26`D. `15` |
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Answer» Correct Answer - c Let n collisions are required for the given conditiaon. Then, `((1)/(2))^(n) xx 2 MeV=0.04 xx 10^(-6) MeV` `2^(n) =(2)/(0.04 xx 10^(6)) =50xx10^(6)` After solving above equation, `n=26. |
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