What is the Orbit of Hydrogen Atom for an Electron at 734 km/s?

AI Thread Summary
The discussion focuses on determining the orbit of a hydrogen atom for an electron traveling at 734 km/s using Bohr's second postulate. The calculations indicate that if the radius is set at 0.0529 nm, it corresponds to the first Bohr orbit (n=1), but for higher energy levels (n>1), the radius increases. Participants suggest that an additional equation is needed to relate the radius to either the principal quantum number (n) or the electron's velocity (v). The need for a balance between centripetal and Coulomb forces is highlighted as a potential solution. Overall, the conversation emphasizes the importance of using the correct equations to find the appropriate orbit for the given speed.
gruba
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Homework Statement


On which orbit of hydrogen atom an electron has the speed of 734 km/s?

Homework Equations


Bohr's second postulate: mvr=nh,m=9.109\cdot 10^{-31}kg,v=734 km/s,h=6.626\cdot 10^{-34} m^{2}kg/s

The Attempt at a Solution


By using the second Bohr's postulate, we get 6686.006\cdot 10^{-28}r=6.626\cdot 10^{-34}n

Question: What is the value of radius of orbit of electron, r?

If we set that r=0.0529 nm, then we get n=0.053 and n should be positive integer.
 
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gruba said:
If we set that r=0.0529 nm
That value is for the first Bohr orbit, ##n=1##. For ##n > 1##, the radius is bigger. You need another equation relating ##r## to ##n## or ##r## to ##v##.
 
DrClaude said:
That value is for the first Bohr orbit, ##n=1##. For ##n > 1##, the radius is bigger. You need another equation relating ##r## to ##n## or ##r## to ##v##.

What is that equation?
 
You can use the equation that balances the centripetal force and the Coulomb force. Don't you have that equation in your notes/textbook?
 
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