Kinetic Energy of a hydrogen electron

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SUMMARY

The kinetic energy of a ground-state electron in a hydrogen atom, as per the Bohr model, can be calculated using the formula KE = 1/2mv^2. The mass of a hydrogen atom is approximately 1.674 x 10^-27 kg, and the orbital speed of the electron is 2190 km/s, which converts to 2190000 m/s. The calculated kinetic energy is 4 x 10^-15 Joules. The discussion highlights the need for the mass of the electron, which is essential for accurate calculations.

PREREQUISITES
  • Understanding of the Bohr model of the atom
  • Familiarity with kinetic energy formula KE = 1/2mv^2
  • Knowledge of unit conversions (e.g., km/s to m/s)
  • Basic understanding of atomic mass and electron mass
NEXT STEPS
  • Research the mass of an electron (9.109 x 10^-31 kg)
  • Explore advanced topics in quantum mechanics related to electron behavior
  • Learn about the implications of the Bohr model in modern physics
  • Investigate other methods for calculating kinetic energy in different contexts
USEFUL FOR

Students studying physics, particularly those focusing on atomic structure and kinetic energy calculations, as well as educators looking for examples of the Bohr model in action.

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Homework Statement



In the Bohr model of the atom, the ground-state electron in hydrogen has an orbital speed of 2190 km/s. What is its kinetic energy?

Homework Equations



KE = 1/2mv^2

The Attempt at a Solution



I'm not sure what I am doing wrong but the program we use to enter homework is saying I'm wrong. So anyway kinetic energy = 1/2*m*v^2. Mass of 1 hydrogen atom = 1.008g/mol*(1mol/6.022*10^23 )*(1kg/1000g) = 1.674*10^-27 g/atom.
The speed is 2190km/s*1000m/km = 2190000m/s therefore
KE = 1/2(1.674*10^-27)(2190000)^2 = 4*10^-15 Joules
 
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It asks for the kinetic energy of the electron. What is the mass of an electron?
 

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