Minimum Kinetic energy of the hydrogen atom

In summary: KE, am I right in saying that the energy is just enough to get the electron in atom 2 to be excited and not impart KE to the atom 2 itself to experience a momentum and therefore minimum KE of atom is...In summary, the minimum kinetic energy of the hydrogen atom such that the hydrogen atom at rest will have its electron in the first excited state after collision is zero.
  • #1
Rahulrj
107
0

Homework Statement


A hydrogen atom collides with another hydrogen atom at rest. If the electrons in both atoms are in the ground state, what is the minimum kinetic energy of the hydrogen atom such that the hydrogen atom at rest will have its electron in the first excited state after collision?

Homework Equations


##E=h \nu##
##E_n=E_1/n^2##

The Attempt at a Solution


I know that the first excited state mean n=2 however, I am not sure how to write the energy conservation here as the question does not mention the energy of the colliding atom after the collision.
 
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  • #2
Rahulrj said:

Homework Equations


##E=h \nu##
Not sure how this is relevant.

Rahulrj said:
the question does not mention the energy of the colliding atom after the collision.

Rahulrj said:
what is the minimum kinetic energy
 
  • #3
DrClaude said:
Not sure how this is relevant.
I don't think I see what you wanted me to see. Need a hint more than that. The only formula I know for KE is the relativistic one ##(\gamma - 1) mc^2## and ##1/2mv^2## but I don't see how to get it from the available info given in the question.
 
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  • #4
Before the collision, where is the energy? After the collision, where can the energy be?
 
  • #5
DrClaude said:
Before the collision, where is the energy? After the collision, where can the energy be?
The colliding atom has the energy before collision and after collision, some energy from the colliding atom is imparted to the atom at rest.
 
  • #6
Rahulrj said:
The colliding atom has the energy before collision and after collision, some energy from the colliding atom is imparted to the atom at rest.
I think it would help if you gave a more detailed answer: what kind of energy has each atom before and after?
 
  • #7
DrClaude said:
I think it would help if you gave a more detailed answer: what kind of energy has each atom before and after?
The colliding atom has kinetic energy and the atom at rest has potential energy. After collision, the kinetic energy of the colliding atom lowers and that amount is compensated by the change in energy of the atom at rest(to kinetic). Since its been given that the electron in the atom at rest excites to n=2, it absorbs photon of equivalent energy?
 
  • #8
Rahulrj said:
The colliding atom has kinetic energy and the atom at rest has potential energy. After collision, the kinetic energy of the colliding atom lowers and that amount is compensated by the change in energy of the atom at rest(to kinetic).
I'm not sure what potential energy you are considering (none is mention in the problem). Let me rephrase what you wrote as (calling atom 2 the atom initially at rest)

before: kinetic energy of atom 1
after: kinetic energy of atom 1 + kinetic energy of atom 2 + electronic energy of atom 2

As you said in the OP, you need to use conservation of energy. As you are to find the minimum energy for which this "after" is possible, what is the minimum value of energy after the collision?

Rahulrj said:
Since its been given that the electron in the atom at rest excites to n=2, it absorbs photon of equivalent energy?
There are no photons involved here.
 
  • #9
DrClaude said:
I'm not sure what potential energy you are considering (none is mention in the problem). Let me rephrase what you wrote as (calling atom 2 the atom initially at rest)

before: kinetic energy of atom 1
after: kinetic energy of atom 1 + kinetic energy of atom 2 + electronic energy of atom 2

As you said in the OP, you need to use conservation of energy. As you are to find the minimum energy for which this "after" is possible, what is the minimum value of energy after the collision?There are no photons involved here.
I only know how to find the electronic energy here and I have not much idea on how to find the KE as I don't know what formula to use there.
 
  • #10
Rahulrj said:
I only know how to find the electronic energy here and I have not much idea on how to find the KE as I don't know what formula to use there.
You already gave the equation for the kinetic energy. But it is not that relevant. Again, what is the minimum kinetic energy you can have after the collision?
 
  • #11
DrClaude said:
You already gave the equation for the kinetic energy. But it is not that relevant. Again, what is the minimum kinetic energy you can have after the collision?
Minimum energy the colliding atom 'can' have after collision is zero? It is possible that all of it can be imparted to atom 2 right?
since the colliding atom has minimum KE, am I right in saying that the energy is just enough to get the electron in atom 2 to be excited and not impart KE to the atom 2 itself to experience a momentum and therefore minimum KE of atom is ##E_1/n^2##?
 
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  • #12
Rahulrj said:
Minimum energy the colliding atom 'can' have after collision is zero? It is possible that all of it can be imparted to atom 2 right?
If the colliding atom comes to rest, what does momentum conservation say about the final speed of the other atom? What energy will that leave for the excitation?
 
  • #13
haruspex said:
If the colliding atom comes to rest, what does momentum conservation say about the final speed of the other atom? What energy will that leave for the excitation?
Using Momentum conservation I can say ##p_1=p_2## so the speeds are same for both? but that doesn't make sense because
If the speeds are same then from energy conservation ##KE_1 = KE'_1+KE_2+E_{electronic}##, ##KE'_1 =0## and ##KE_1-KE_2 =E_{electronic}## (1 is for colliding atom and 2 is for the other atom) the two kinetic energies have to be zero right?
 
  • #14
Rahulrj said:
Using Momentum conservation I can say ##p_1=p_2## so the speeds are same for both?
Define p1 and p2 there. If you mean total momentum before and after then yes, that is conservation of momentum, but it does not mean they would have the same final speed. (They just might, though.)
Rahulrj said:
##KE_1 = KE'_1+KE_2+E_{electronic}##
Right.
Rahulrj said:
##KE'_1 =0##
No, why?
 
  • #15
haruspex said:
Define p1 and p2 there. If you mean total momentum before and after then yes, that is conservation of momentum, but it does not mean they would have the same final speed. (They just might, though.)

Right.

No, why?
##KE'_1 = 0## is the KE of colliding atom after collision which you said it comes to rest and that means KE is zero right?
##p_1## is the momentum of colliding atom before collision and ##p_2## is the momentum of the atom at rest after collision
 
  • #16
Rahulrj said:
colliding atom after collision which you said it comes to rest
No, I did not say it comes to rest. I asked, IF it comes to rest (as you had assumed), what would conservation of momentum tell you about the final speed of the other atom? I asked this inorder to demonstrate that it certainly does not come to rest.

You have an equation for conservation of energy and another for conservation of momentum. They involve two unknown final velocities and one unknown initial velocity. Two equations with three unknowns leaves one degree of freedom. As you vary one velocity, the other two velocities vary in consequence. Your task is to find how to minimise the unknown initial velocity.
 
  • #17
I still didn't get how you came to the conclusion that the colliding atom does not come to rest. According to my thinking, to minimize unknown initial velocity just enough to get the electron in atom 2 to be excited means the kinetic energy of the colliding atom after collision has to be zero as the question is asking only for a minimum KE.
 
  • #18
Rahulrj said:
I still didn't get how you came to the conclusion that the colliding atom does not come to rest
If the incoming atom comes to rest, conservation of momentum says the other atom, being of the same mass, must acquire the speed that it lost. The would mean that it also acquires all the KE, leaving no energy for the excitation.

Please, write out the equations as I suggested.
 
  • #19
haruspex said:
If the incoming atom comes to rest, conservation of momentum says the other atom, being of the same mass, must acquire the speed that it lost. The would mean that it also acquires all the KE, leaving no energy for the excitation.
Okay, so writing down the equations
##mv_{1i} = mv_{1f}+mv_{2f}##
## 1/2mv_{1i}^2 = 1/2mv_{1f}^2+1/2mv_{2f}^2+E_{electronic}## substituting ##v_{1i}## from momentum equation to energy equation
##mv_{1f}v_{2f}= E_{electronic}##
Not sure what I should be looking for in these equations to get minimum velocity.
 
  • #20
Rahulrj said:
Okay, so writing down the equations
##mv_{1i} = mv_{1f}+mv_{2f}##
## 1/2mv_{1i}^2 = 1/2mv_{1f}^2+1/2mv_{2f}^2+E_{electronic}## substituting ##v_{1i}## from momentum equation to energy equation
##mv_{1f}v_{2f}= E_{electronic}##
Not sure what I should be looking for in these equations to get minimum velocity.
You know the product of two numbers, when is their sum minimum? (Think of the relation between the geometric mean and the arithmetic mean.)
 
  • #21
Rahulrj said:
Not sure what I should be looking for in these equations to get minimum velocity.
You can use ehild's excellent hint, or use your last equation to substitute for v2f in your first equation. That would give you one equation with two unknowns, v1i and v1f. How do you minimise one variable with respect to another?
 
  • #22
haruspex said:
You can use ehild's excellent hint, or use your last equation to substitute for v2f in your first equation. That would give you one equation with two unknowns, v1i and v1f. How do you minimise one variable with respect to another?
I am not very familiar solving using AM and GM concepts, it has been a while but I would love to know how it can be done. So I choose to go with the substitution method and the equation I get is ##v_{1i}=v_{1f}+E_{electronic}/mv_{1f}##. So now do I need to take a derivative w.r.t ##v_{1f}## to find minimum?
 
  • #23
Rahulrj said:
I am not very familiar solving using AM and GM concepts, it has been a while but I would love to know how it can be done. So I choose to go with the substitution method and the equation I get is ##v_{1i}=v_{1f}+E_{electronic}/mv_{1f}##. So now do I need to take a derivative w.r.t ##v_{1f}## to find minimum?
Yes.
 

Related to Minimum Kinetic energy of the hydrogen atom

1. What is the minimum kinetic energy of a hydrogen atom?

The minimum kinetic energy of a hydrogen atom refers to the minimum amount of energy required for the atom to be in motion. This energy is known as the zero-point energy and has a value of 1/2 * Planck's constant * frequency of the atom.

2. How is the minimum kinetic energy of a hydrogen atom calculated?

The minimum kinetic energy of a hydrogen atom can be calculated using the formula E = 1/2 * mv^2, where m is the mass of the atom and v is the velocity. However, for a hydrogen atom, the minimum kinetic energy is determined by the zero-point energy, as mentioned in the previous question.

3. Why is the minimum kinetic energy of a hydrogen atom important?

The minimum kinetic energy of a hydrogen atom is important because it is the lowest possible energy state for the atom. This energy state is crucial for understanding the behavior and properties of atoms, as it is the starting point for all other energy states.

4. How does the minimum kinetic energy of a hydrogen atom relate to its electronic structure?

The minimum kinetic energy of a hydrogen atom is closely related to its electronic structure, as it is determined by the energy levels of the atom's electrons. In the lowest energy state, the electrons are in their ground state and have the lowest possible energy levels.

5. Can the minimum kinetic energy of a hydrogen atom be observed in experiments?

No, the minimum kinetic energy of a hydrogen atom cannot be observed directly in experiments. However, its effects can be seen in various phenomena such as the stability of atoms and the emission and absorption of light. The zero-point energy has also been indirectly measured in experiments involving quantum mechanics.

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