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

Click For 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
Messages
203
Reaction score
1

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.
 
Physics news on Phys.org
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?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
2
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 5 ·
Replies
5
Views
4K