Center of Mass, Momentum & Collision

JEE Main level, numerical answer, medium

105 of 130

Question 105 of 130

A particle of mass 200MeVc−2200MeV{c}^{-2} collides with a hydrogen atom at rest. Soon after the collision, the particle comes to rest, and the atom recoils and goes to its first excited state. The initial kinetic energy of the particle (in eVeV) is N4.\dfrac{N}{4}. The value of NN is: (Given the mass of the hydrogen atom to be 1GeVc−21GeV{c}^{-2}).........

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Answer: 51

Given, M=200MeVc−2M=200MeV{c}^{-2} and m=1GeVc−2m=1GeV{c}^{-2}, now use the concept of conservation of linear momentum, Mv0=mv⇒v=Mv0m...(1)M{v}_{0}=mv\Rightarrow v=\dfrac{M{v}_{0}}{m}...\left(1\right).
Now use the conservation of energy,
12M(v0)2=12m(v)2+∆E..(2)\dfrac{1}{2}M{\left({v}_{0}\right)}^{2}=\dfrac{1}{2}m{\left(v\right)}^{2}+∆E..\left(2\right), here ∆E=13.6eV∆E=13.6eV, for the hydrogen atom in the ground state, now from equation, \left(1\right)&\left(2\right), 12M(v0)2=14(3∆E×108)eV=N4\dfrac{1}{2}M{\left({v}_{0}\right)}^{2}=\dfrac{1}{4}\left(3∆E\times \dfrac{10}{8}\right)eV=\dfrac{N}{4},
⇒N4=514eV\Rightarrow \dfrac{N}{4}=\dfrac{51}{4}eV⇒N=51\Rightarrow N=51.

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