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A neutron moving with speed v makes a head on collision with a hydrogen atom in ground state kept at rest. The minimum kinetic energy of the neutron for which inelastic collision takes place is
- a)10.2 eV
- b)20.4 eV
- c)12.1 eV
- d)16.8 eV
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A neutron moving with speed v makes a head on collision with a hydroge...
Solution:
In an inelastic collision, the kinetic energy of the system is not conserved. The neutron will transfer some of its kinetic energy to the hydrogen atom, exciting the atom to a higher energy level.
Let mass of neutron be mn and mass of hydrogen atom be mh.
Let u be the velocity of hydrogen atom after the collision and v be the velocity of neutron after the collision.
By conservation of momentum:
mnv = mhu
By conservation of energy:
(1/2)mnv^2 = (1/2)mhu^2 + E, where E is the energy transferred to the hydrogen atom causing its excitation.
The minimum kinetic energy of the neutron for which inelastic collision takes place is when all the kinetic energy is transferred to the hydrogen atom for its excitation. In this case, u = 0.
mnv = mhu = √(2E/mh)
(1/2)mnv^2 = E
Substituting the values, we get:
mnv^2 = 2E
v^2 = (2E/mn)
v = √(2E/mn)
Given, mass of neutron, mn = 1.67 x 10^-27 kg
Mass of hydrogen atom, mh = 1.67 x 10^-27 kg
Speed of neutron, v = 4.4 x 10^5 m/s
We need to find the minimum kinetic energy, E.
E = (1/2)mnv^2
E = (1/2) x 1.67 x 10^-27 x (4.4 x 10^5)^2
E = 20.4 eV
Therefore, the minimum kinetic energy of the neutron for which inelastic collision takes place is 20.4 eV.
Hence, option B is the correct answer.
In an inelastic collision, the kinetic energy of the system is not conserved. The neutron will transfer some of its kinetic energy to the hydrogen atom, exciting the atom to a higher energy level.
Let mass of neutron be mn and mass of hydrogen atom be mh.
Let u be the velocity of hydrogen atom after the collision and v be the velocity of neutron after the collision.
By conservation of momentum:
mnv = mhu
By conservation of energy:
(1/2)mnv^2 = (1/2)mhu^2 + E, where E is the energy transferred to the hydrogen atom causing its excitation.
The minimum kinetic energy of the neutron for which inelastic collision takes place is when all the kinetic energy is transferred to the hydrogen atom for its excitation. In this case, u = 0.
mnv = mhu = √(2E/mh)
(1/2)mnv^2 = E
Substituting the values, we get:
mnv^2 = 2E
v^2 = (2E/mn)
v = √(2E/mn)
Given, mass of neutron, mn = 1.67 x 10^-27 kg
Mass of hydrogen atom, mh = 1.67 x 10^-27 kg
Speed of neutron, v = 4.4 x 10^5 m/s
We need to find the minimum kinetic energy, E.
E = (1/2)mnv^2
E = (1/2) x 1.67 x 10^-27 x (4.4 x 10^5)^2
E = 20.4 eV
Therefore, the minimum kinetic energy of the neutron for which inelastic collision takes place is 20.4 eV.
Hence, option B is the correct answer.
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Community Answer
A neutron moving with speed v makes a head on collision with a hydroge...
Let speed of neutr on before collision = V Speed of neutron after collision = V1
Speed of proton or hydrogen atom after collision = V2
Energy of excitation = DE From the law of conservation of linear momentum, mv = mv1 + mv2 ...(1)
And for law of conservation of energy,
...(2)
From squaring eq. (i), we get
...(3)
From squaring eq. (ii), we get
...(4)
From eqn (3) & (4)
As, v1 – v2 must be real,
⇒
The minimum energy that can be absorbed by the hydrogen atom in the ground state to go into the excited state is 10.2 eV. Therefore, the maximum kinetic energy needed is
2 x 10.2 = 20.4 eV
Speed of proton or hydrogen atom after collision = V2
Energy of excitation = DE From the law of conservation of linear momentum, mv = mv1 + mv2 ...(1)
And for law of conservation of energy,
...(2)From squaring eq. (i), we get
...(3)From squaring eq. (ii), we get
...(4)From eqn (3) & (4)
As, v1 – v2 must be real,
⇒
The minimum energy that can be absorbed by the hydrogen atom in the ground state to go into the excited state is 10.2 eV. Therefore, the maximum kinetic energy needed is
2 x 10.2 = 20.4 eVAttention JEE Students!
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A neutron moving with speed v makes a head on collision with a hydrogen atom in ground state kept at rest. The minimum kinetic energy of the neutron for which inelastic collision takes place isa)10.2 eVb)20.4 eVc)12.1 eVd)16.8 eVCorrect answer is option 'B'. Can you explain this answer?
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