Question

An alpha particle and a proton are accelerated through the same potential difference. Calculate the ratio of linear momentum acquired by the two particles.

Answer 1

ratio of their linear momentum acquired by the two particles is 2√2 : 1

let a alpha particle and a proton are accelerated through the same potential difference V.

then, work done = kinetic energy

for alpha particle : charge on each alpha particle, q₁ = + 2e [ where e = 1.6 × 10^-19 C]

mass of alpha particle, m₁ = 4m [ where m is mass of a hydrogen atom ]

so, q₁V = P₁²/2m₁

⇒P₁ = √{2q₁Vm₁}

= √{2(2e)V(4m)

= √{16eVm}............(1)

for proton : charge on each proton, q₂ = +1e

mass of proton, m₂ = m

so, q₂V = P₂²/2m₂

⇒P₂= √{2q₂Vm₂}

= √{2(1e)V(m)}

= √{2eVm}...........(2)

from equations (1) and (2) we get,

P₁/P₂ = √{16eVm/2eVm} = 2√2/1

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Answer 2

$\underline{ \huge \boxed{ \bold{ \mathfrak{ \purple{Answer}}}}}$

Given :

An alpha particle and a proton are accelerated through the same potential difference.

To Find :

The ratio of linear momentum acquid by both paticles.

Formula :

Relation between linear momentum(P) and Potential difference(V) is given by...

$\: \: \: \: \dag \: \: \underline{ \boxed{ \bold{ \huge{ \rm{ \pink{p = \sqrt{2mqV}}}}}}} \: \: \dag$

Calculation :

We know that ,

1. mass of alpha particle is four times of mass of proton.
2. Charge on alpha particle is two times of charge on ptoton.

As per formula it is clear that...

$\implies \rm \: \red{P \propto{ \sqrt{mq} }} \\ \\ \implies \rm \: \frac{P_{alpha}}{P_{proton}} = \sqrt{ \frac{(mq)_{alpha}}{(mq)_{proton}} } = \sqrt{8} \\ \\ \therefore \: \underline{ \boxed{ \bold{ \rm{ \orange{P_{alpha} :P_{proton} = 2\sqrt{2} : 1}}}}}$