A neutron moving with a speed ‘v’ makes a head on collision with a stationary hydrogen atom in ground state. The minimum kinetic energy of the neutron for which inelastic collision will take place is:
A. $20.4eV$
B. $10.2eV$
C. $12.1eV$
D. $16.8eV$
Answer
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Hint: We know that, the conservation of linear momentum states that the net external force acting on a system of bodies is zero, then the momentum of the system remains constant. Law of conservation of energy is defined as that the energy can neither be created nor be destroyed, it may be transformed from one form to another. Using this concept, we can find the solution of the given question.
Complete answer:
It is given that the speed of the neutron before collision is ‘V’.
Let us assume that the speed of the neutron after collision is ‘${V_1}$’
Speed of the hydrogen atom after collision is assumed to be ‘${V_2}$’.
Excitation energy $ = \Delta E$
According to the law of conservation of linear momentum,
$mv = m{v_1} + m{v_2}$ -(1)
According to the law of conservation of energy,
$\dfrac{1}{2}m{v^2} = \dfrac{1}{2}m{v_1}^2 + \dfrac{1}{2}m{v_2}^2$ -(2)
Squaring the equation (1) we will get,
${v^2} = {v_1}^2 + {v_2}^2 + 2{v_1}{v_2}$ -(3)
Squaring the equation (2) we will get,
${v^2} = {v_1}^2 + {v_2}^2 + \dfrac{{2\Delta E}}{m}$ -(4)
Equating the equation (1) and equation (2) we will get,
$2{v_1}{v_2} = \dfrac{{2\Delta E}}{m}$
$ \Rightarrow {\left( {{v_1} - {v_2}} \right)^2} = {\left( {{v_1} + {v_2}} \right)^2} - 4{v_1}{v_2} = {v^2} - \dfrac{{4\Delta E}}{m}$
Since, ${v_1} - {v_2}$ must be real
So, ${v^2} - \dfrac{{4\Delta E}}{m} \geqslant 0$
$ \Rightarrow \dfrac{1}{2}m{v^2} \geqslant 2\Delta E$
The minimum energy which can be absorbed by a hydrogen atom in the ground state to jump into an excited state is $10.2eV$. Thus, the minimum kinetic energy of the neutron for which inelastic collision will take place is,
$\dfrac{1}{2}m{v_{\min }}^2 = 2 \times 10.2 = 20.4eV$
Hence, option (A) is the correct answer.
Note:
A collision is defined as the event in which two or more objects exert forces on each other for a very short interval of time. It is further classified in two types, inelastic collision and elastic collision. An inelastic collision is defined as the collision which occurs between two objects in which some of the energy is lost. In this collision the momentum is conserved but the kinetic energy is not conserved.
Complete answer:
It is given that the speed of the neutron before collision is ‘V’.
Let us assume that the speed of the neutron after collision is ‘${V_1}$’
Speed of the hydrogen atom after collision is assumed to be ‘${V_2}$’.
Excitation energy $ = \Delta E$
According to the law of conservation of linear momentum,
$mv = m{v_1} + m{v_2}$ -(1)
According to the law of conservation of energy,
$\dfrac{1}{2}m{v^2} = \dfrac{1}{2}m{v_1}^2 + \dfrac{1}{2}m{v_2}^2$ -(2)
Squaring the equation (1) we will get,
${v^2} = {v_1}^2 + {v_2}^2 + 2{v_1}{v_2}$ -(3)
Squaring the equation (2) we will get,
${v^2} = {v_1}^2 + {v_2}^2 + \dfrac{{2\Delta E}}{m}$ -(4)
Equating the equation (1) and equation (2) we will get,
$2{v_1}{v_2} = \dfrac{{2\Delta E}}{m}$
$ \Rightarrow {\left( {{v_1} - {v_2}} \right)^2} = {\left( {{v_1} + {v_2}} \right)^2} - 4{v_1}{v_2} = {v^2} - \dfrac{{4\Delta E}}{m}$
Since, ${v_1} - {v_2}$ must be real
So, ${v^2} - \dfrac{{4\Delta E}}{m} \geqslant 0$
$ \Rightarrow \dfrac{1}{2}m{v^2} \geqslant 2\Delta E$
The minimum energy which can be absorbed by a hydrogen atom in the ground state to jump into an excited state is $10.2eV$. Thus, the minimum kinetic energy of the neutron for which inelastic collision will take place is,
$\dfrac{1}{2}m{v_{\min }}^2 = 2 \times 10.2 = 20.4eV$
Hence, option (A) is the correct answer.
Note:
A collision is defined as the event in which two or more objects exert forces on each other for a very short interval of time. It is further classified in two types, inelastic collision and elastic collision. An inelastic collision is defined as the collision which occurs between two objects in which some of the energy is lost. In this collision the momentum is conserved but the kinetic energy is not conserved.
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