Question

The alpha particles and a proton are accelerated through the same potential difference.calculate the ratio of linear momenta acquired by the two.

Answer 1

Mass of proton = Mp = 1u
Mass of alpha particle = Ma = 4u
charge on proton = Qp = + 1e
Charge on alpha particle =+ 2e

Now, we should use the formula,
$\bold{P = \sqrt{2qVm}}$
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 :
$\bold{P_{proton}=\sqrt{2\times1e\times V\times1u}}$
similarly, liner momentum of alpha particle :
$\bold{P_{\alpha-particle}=\sqrt{2\times2e\times V\times4u}}$

ratio of momentum of proton to alpha particle will be $\bold{\frac{P_{proton}}{P_{\alpha-particle}}=\sqrt{\frac{1}{8}}}=\frac{1}{2\sqrt{2}}$
ratio of momentum of alpha particle to proton is 2√2:1

Answer 2

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