1. The average KE and temperature in Kelvin of the molecules of a gas are related by the equation KE = 3/2 KT where k is the Boltzmann constant 1.38 x 10 m² kg s². The diagram shows the energy levels for a Hydrogen atom. Energy/eV 0.00 -1.51 3.39 13.58 Use this information to show that Hydrogen at room temperature will not emit light. 2. When hydrogen burns in oxygen 241.8 kJ of energy are released per mole. Show that this reaction can produce light.
1. The average KE and temperature in Kelvin of the molecules of a gas are related by the equation KE = 3/2 KT where k is the Boltzmann constant 1.38 x 10 m² kg s². The diagram shows the energy levels for a Hydrogen atom. Energy/eV 0.00 -1.51 3.39 13.58 Use this information to show that Hydrogen at room temperature will not emit light. 2. When hydrogen burns in oxygen 241.8 kJ of energy are released per mole. Show that this reaction can produce light.
Chapter10: Atomic Physics
Section: Chapter Questions
Problem 12P
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
Transcribed Image Text:1. The average KE and temperature in Kelvin of the molecules of a gas are related by the
equation KE = 3/2 KT where k is the Boltzmann constant 1.38 x 10 m² kg s².
The diagram shows the energy levels for a Hydrogen atom.
Energy/eV
0.00
-1.51
3.39
13.58
Use this information to show that Hydrogen at room temperature will not emit light.
2. When hydrogen burns in oxygen 241.8 kJ of energy are released per mole. Show that this
reaction can produce light.
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