Physics Angular Momentum of Electron?

Question

In the Bohr model of the hydrogen atom, the electron in the n = 10 level moves in a circular orbit of radius 5.29 x10-9 m around the proton. Assume the orbital angular momentum of the electron is equal to 10h/2.

Okay, I found in a book that h = 1.054x10^-34 kgm^2/s

I used the equation L = Iw
I = mr^2
and then plugged the values in to get w.

I don't understand what the n =10 means so my answer is wrong. Please help!!!

(a) Calculate the orbital speed of the electron.
____m/s
(b) Calculate the kinetic energy of the electron.
____ J
(c) Calculate the angular frequency of the electron's motion.
____ rad/s

Answer 1

Orbital angular momentum is

L^2 = l (l+1) (h-cross)^2

l = angular momentum quantum number, h-cross = h /2pi
for n=10
-----------------------
Angular momentum L = mv *cros* r that is moment of momentum

as per Bohr theory L = mvr = n h/2pi

for n = 10, L = 10 h/2 Pi (you have taken it 10h/2 instead

calculate by mvr = 10*h/2*3.14

v = L / m r = 10*h / [2*3.14 * 9.1*10^-31 * 5.29 *10^-9] m/s

you know the mass of electron calculate v

KE = 0.5 m v^2

angular freq = w = v / r rad/sec

Answer 2

I can't give you the math here to help you as space doesn't permit. BUT, if you consider the wave state analysis of the electron pair (that thing that keeps hydrogen atoms jumping around from one to another) and the spin on them, you will be headed in the right direction. Hope this helps.