Ionization energy and speed of electron

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SUMMARY

The discussion centers on calculating the speed of an electron required to eject another electron from a neon atom, which has a first ionization energy of 2080 kJ/mol. Two solutions were proposed: the first calculated the speed using the mass of a neon atom, resulting in 1.436 x 104 m/s, while the second used the mass of an electron, yielding 2.754 x 106 m/s. The correct approach is the second solution, as it accurately reflects the kinetic energy conversion necessary for ionization. The context of neon gas ionization in discharge tubes was also clarified.

PREREQUISITES
  • Understanding of ionization energy and its measurement in kJ/mol
  • Familiarity with Avogadro's number and its application in atomic calculations
  • Knowledge of kinetic energy equations, specifically E = (1/2)mv2
  • Basic principles of gas discharge tubes and electron behavior in ionization
NEXT STEPS
  • Research the concept of ionization energy in different elements and its implications
  • Learn about the behavior of electrons in gas discharge tubes and their role in light emission
  • Explore advanced calculations involving kinetic energy and particle physics
  • Study the differences in ionization energies across noble gases and their applications in lighting
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Physicists, chemistry students, electrical engineers, and anyone interested in the principles of gas ionization and electron dynamics in discharge tubes.

lkh1986
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A neon sign is a gas discharge tube in which electrons traveling from the cathode to anode collide with neon atoms in dicharge tube and knocks electrons off of them. As electrons return to the neon ions and drop to lower energy levels, light is given off. How fast would an electron have to be moving to eject an electron from an atom of neon,which has a first ionization energy of 2080KJ per mol.

OK, I got 2 set of solutions, buI don't know which one of them is the correct one, maybe neither is correct, too.

For solution number 1, I calculate the ionization energy for 1 atom of neon, by dividing 2080 by Avogadro number. Then, I get E. I obtain the mass of 1 atom of neon, by dividing its molar mass by Avogadro number, too. Using E= (1/2)(mv^2), i get the value of v=1.436x10^4 m/s.

For solution number 2, i do the same thing, except for the value of m, I substitute in the m for an electron. Finally, I get v=2.754x10^6 m/s.

Which one is the correct solution? Or neither? Thanks. :smile:
 
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One solution is correct, the other is incorrect. Which do you think is which and why?
 
Last edited:
I think the solution which I substitute the value of m equals the mass of an electron seems correct. My understanding of the situation of this problem is as such: the kinetic energy of the high-speed moving electron has been "changed" to the energy that is required to knock out an electron from a neon atom, which is later become te light energy.

May I know where the electron that are being accelerated comes from? Is it from the neon atom itself? Or it is "already there" in the discharge tube? Thanks.
 
lkh1986 said:
I think the solution which I substitute the value of m equals the mass of an electron seems correct. My understanding of the situation of this problem is as such: the kinetic energy of the high-speed moving electron has been "changed" to the energy that is required to knock out an electron from a neon atom, which is later become te light energy.
Yes, the latter answer is correct.
May I know where the electron that are being accelerated comes from? Is it from the neon atom itself? Or it is "already there" in the discharge tube? Thanks.
A brief introduction from howstuffworks.com;
Howstuffworks said:
The idea behind a neon light is simple. Inside the glass tube there is a gas like neon, argon or krypton at low pressure. At both ends of the tube there are metal electrodes. When you apply a high voltage to the electrodes, the neon gas ionizes, and electrons flow through the gas. These electrons excite the neon atoms and cause them to emit light that we can see. Neon emits red light when energized in this way. Other gases emit other colors.
 

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