Do i have to answer the question and still have the offer on the table
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 ×
/ ( 250 × π × 20 )
1300 ×
= (0.286)² × W ×
/ ( 250 × π × 20 )
solve it and we get W
W = 249.56 ×
so elastic partial width is 2.49 eV
Answer:
2000 ohms
Explanation:
Resisters in series just add.
Rt = R1 + R2 + R3
R1 = 650 ohm
R2 = 350 ohm
R3 = 1000 ohm
Rt = 650 + 350 + 1000
Rt = 2000 ohms.
What equation?
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