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Solved: In the Bohr model of the hydrogen atom, an
Chapter 8, Problem 7E(choose chapter or problem)
In the Bohr model of the hydrogen atom, an electron (mass \(m=9.1 \times 10^{-31} \mathrm{~kg}\) ) orbits a proton at a distance of \(5.3 \times 10^{-11} \mathrm{~m}\). The proton pulls on the electron with an electric force of \(8.2 \times 10^{-8} \mathrm{~N}\). How many revolutions per second does the electron make?
Equation Transcription:
Text Transcription:
m= 9.1 X 10^{-31} kg
5.3 X 10^{-11} m
8.2 X 10^{-8} N
Questions & Answers
QUESTION:
In the Bohr model of the hydrogen atom, an electron (mass \(m=9.1 \times 10^{-31} \mathrm{~kg}\) ) orbits a proton at a distance of \(5.3 \times 10^{-11} \mathrm{~m}\). The proton pulls on the electron with an electric force of \(8.2 \times 10^{-8} \mathrm{~N}\). How many revolutions per second does the electron make?
Equation Transcription:
Text Transcription:
m= 9.1 X 10^{-31} kg
5.3 X 10^{-11} m
8.2 X 10^{-8} N
ANSWER:
Step 1 of 2
Here we have to calculate the number of revolutions per second the electron makes around the proton.
The mass of the electron is,
\(m_{e}=9.1 \times 10^{-31} \mathrm{~kg}\)
The distance between the proton and electron is,
\(r=5.3 \times 10^{-11} m\)
The electric force between them is,
\(F_{e l}=8.2 \times 10^{-8} N .\)