seto6
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this question is from my previous assignment which was due 2 weeks back. just revising for mid-term not sure how to solve it. if someone explains it thoroughly it would be grate
An alpha particle is a nucleus of Helium. It has twice the charge and four times the mass of the proton.
A proton and an alpha particle headed directly toward each other, had each initial speed of 3.9×10−3 c when they were far away.
Here, as is customary when describing processes involving nuclear targets, the speed is expressed as a fraction of c, the speed of light.
What is the distance of closest approach between the proton and the alpha particle?
Hint: There are two conserved quantities. Make use of both.
U=Eqd or KqQ/r
K=1/2*(mv^2)
.5m(v_i)^2 + Eq(d_i)= .5m(v_f)^2 + Eq(d_f)
where (V_f)=0 then i solved for (d_f)
and got the 1.87*10^-27 but the answer should be d_f=1.3*10^-13
could some 1 explain this to me and the reasoning! please
Homework Statement
An alpha particle is a nucleus of Helium. It has twice the charge and four times the mass of the proton.
A proton and an alpha particle headed directly toward each other, had each initial speed of 3.9×10−3 c when they were far away.
Here, as is customary when describing processes involving nuclear targets, the speed is expressed as a fraction of c, the speed of light.
What is the distance of closest approach between the proton and the alpha particle?
Hint: There are two conserved quantities. Make use of both.
Homework Equations
U=Eqd or KqQ/r
K=1/2*(mv^2)
The Attempt at a Solution
.5m(v_i)^2 + Eq(d_i)= .5m(v_f)^2 + Eq(d_f)
where (V_f)=0 then i solved for (d_f)
and got the 1.87*10^-27 but the answer should be d_f=1.3*10^-13
could some 1 explain this to me and the reasoning! please
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