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
1 answer:
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
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and the gravitational potential energy of the rock has not changed, since the rock has returned into its initial position, it means that the speed of the rock should be the same)
This means that we can only analyze the final part of the motion, the one in which the rock falls into the 10 m hole. Since it is a free fall motion, we can find the final speed by using
where
u = 21 m/s is the initial speed of the rock as it enters the hole
g = 9.8 m/s^2 is the acceleration due to gravity
d = 10 m is the depth of the hole
Substituting,
(b) 4.72 s
The vertical position of the rock at time t is given by
where
is the initial vertical velocity
Substituting y(t)=-10 m, we can then solve the equation for t to find the time at which the rock reaches the bottom of the hole:
which has two solutions:
t = -0.43 s --> negative, so we discard it
t = 4.72 s --> this is our solution