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

Answer:

elastic partial width is 2.49 eV

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