Question

What is the speed of an e- in the n=4 Bohr orbit?

Answer 1

The electron velocity in the nth Bohr orbit is given by:\[v=\sqrt{\frac{ 2∙13.6Z ^{2} }{ m _{e}n ^{2} }}\]where me is the mass of an electron, 9.11*10^-31kg. Note that under the radical you get units of eV/kg, so you'll need to convert 13.6eV to joules before taking the square root.

Answer 2

Do you know what this is? :
Vn = (2πke^2)/ (nh)

Answer 3

Yes, it's the same as mine but expressed slightly differently. 13.6eV is derived from the following constants:\[13.6eV=\frac{ \left( k _{e} e ^{2}\right) ^{2}m _{e}}{ 2ħ^{2} }=R _{E}\]where ke is Coulomb's constant; e is the electron charge; and ħ is defined as follows:\[ħ=\frac{ h }{ 2\pi }\]where h is Planck's constant. Now if you substitute my expression for RE in place of 13.6 in my equation above for v and then simplify it, you'll get the equation that you've just shown.

Answer 4

Note that Planck's constant can be expressed in J∙s or in eV∙s, so you have to choose the correct value. Note that when RE=13.6eV, Planck's constant is in terms of eV∙s.

Answer 5

so ke^2 is actually (h/2pi)(1.6*10^-19)^2

Answer 6

No ke is Coulomb's constant which is 8.988*10^9 N-m^2/C^2.

Answer 7

ohhhhhh. I usually write that as k=9*10^9. Let me just try to do this problem now.

Answer 8

Nevermind. My result is in the 10^53s

Answer 9

Oh wait. I got it. k=9*10^9 and e=1.6*10^-19

Answer 10

Note that I think in your expression k is what I would call "k sub e" and e is the electron charge. When I type "ke" I mean "k sub e".

Answer 11

Your answer shouldn't be in seconds. When I work it out I get v=5.47*10^5 m/s

Answer 12

I get 5.45869 *10^5 approx

Answer 13

Oh well. I'l off.

Answer 14

That's it.