Question

Neutrons incident on a heavy nucleus with spin J 0 show a resonance at an incident energy ER = 250 eV in the total cross-section with a peak magnitude of 1300 barns, the observed width or the peak being-20 ev. Calculate the elastic partial width of the resonance.

Answer 1

elastic partial width is 2.49 eV

Step-by-step explanation:

given data

ER E = 250 eV

spin J = 0

cross-section magnitude σ = 1300 barns

peak P = 20ev

to find out

elastic partial width W

solution

we know here that

σ = λ²× W / ( E × π × P ) ...................1

put here all value

σ = (0.286)² × W ×
`10^(-16)`/ ( 250 × π × 20 )

1300 ×
`10^(-24)` = (0.286)² × W ×
`10^(-16)`/ ( 250 × π × 20 )

solve it and we get W

W = 249.56 ×
`10^(-2)`

so elastic partial width is 2.49 eV