The alpha particles and a proton are accelerated through the same potential difference.calculate the ratio of linear momenta acquired by the two.
The alpha particles and a proton are accelerated through the same potential difference.calculate the ratio of linear momenta acquired by the two.
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Verified answer
Mass of proton = Mp = 1uMass of alpha particle = Ma = 4u
charge on proton = Qp = + 1e
Charge on alpha particle =+ 2e
Now, we should use the formula,
where P is liner momentum of body
q is charge on body , V is potential difference and m is mass of body
so, liner momentum of proton :
similarly, liner momentum of alpha particle :
ratio of momentum of proton to alpha particle will be
ratio of momentum of alpha particle to proton is 2√2:1
Let's recall a couple of important properties of proton and an alpha particle
mass of proton = 1u and mass of alpha particle=4u
charge on proton=e and charge on alpha particle=2e
now, for charged particle moving in a uniform potential field the following relation holds true
the kinetic energy acquired by then would be equal to the electric potential energy, thus
(1/2)mv2 = qV
or momentum
p = mv = (2qVm)1/2
here m and v are mass and velocity of the particle and q and V are the charge of the particle and the electric potential experienced.
so, for a proton the momentum will be
p1 = (2q1Vm1)1/2 = [2(e)V.(1u)]1/2
similarly for an alpha particle
p2 = (2q2Vm2)1/2 = [2.(2e)V.(4u)]1/2
so, the ratio will be
p1/p2 = [2(e)V.(1u)]1/2 / [2.(2e)V.(4u)]1/2
or
p1/p2 = 1/8