Respuesta :
Using the theories of collision, we got that 13.36% is the the kinetic energy of the alpha particle is transferred to the uranium nucleus if we assume the collision is one dimensional.
So, we know very well that when a collision is elastic, it means law of conservation of momentum and law of conservation of energy is applicable.
Since, it is given that collision is one dimensional, therefore we need to conserve momentum only in x-direction
So, suppose alpha particle is moving with velocity v₁, and uranium is initially at rest so, its velocity is zero.
Now, after the collision, suppose alpha particle moves with velocity(v₂) and uranium nucleus moving with vecloity(v₃)
So, according to law of conservation of momentum,
Intial momentum of the system=final momentum of the system after collision
∴mass of alpha×v₁+mass of uranium×0=mass of alpha×v₂+mass of uranium × v₃
=>4×v₁+0=4×v₂+235×v₃
=>4(v₁-v₂)=235v₃
=>4/235=[v₃/(v₁-v₂)]----------------------(1)
Now, applying law of conservation of energy,
(1/2)×mass of alpha)×v₁²=[([1/2)×mass of alpha×v₂²]+[(1/2)×mass of uranium×v₃²]
=>[(1/2)×4×v₁²)=[ [ (1/2) × 4×v₂²] + [(1/2)×235×v₃²]
=>4v₁²=4v₂²+235v₃²
=>4v₁²-4v₂²=235v₃²
=>4(v₁²-v₂²)=235v₃²
=>(4/235)=[(v₃²)/((v₁²-v₂²)]----------------(2)
We know that a²-b²=(a-b) .(a + b)
After applying this and comparing both of the equations, we get
=>[(v₃²)/((v₁²-v₂²)]=[v₃/(v₁-v₂)]
=>[(v₃²)/[((v₁-v₂)×(v₁+v₂)]]=[v₃/(v₁-v₂)]
=>[v₃/(v₁+v₂)]=1
=>v₃ = v₁+v₂-----------------(eq3)
Now, it is given that v₁=10m/sec
So, putting this, we get
v₃ = 10+v₂
Putting this v₃ in eq1,we get
=>(4/235)=(10+v₂)/(10-v₂)
=>4×(10-v₂)=235×(10+v₂)
=>40-4v₂=2350+235v₂
=>239v₂=-2310
=>v₂=(2310/239)
=>v₂=9.66m/sec
Now kinetic energy of alpha particle before collision=(1/2)×4×10²=200Joule
Now, kinetic energy of alpha particle after collision=(1/2)×4×(9.66)²=186.63Joule
So, net energy which is transferred to uranium is =200-186.63=13.36Joule.
Hence, an alpha particle (4he) undergoes an elastic collision with a stationary uranium nucleus (235u). the percent of the kinetic energy of the alpha particle which is transferred to the uranium nucleus is 13.36 Joule or 13.36%
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(Complete question) is:
an alpha particle (4he) moving with velocity 10m/sec undergoes an elastic collision with a stationary uranium nucleus (235u). what percent of the kinetic energy of the alpha particle is transferred to the uranium nucleus? assume the collision is one dimensional.
