PDA

View Full Version : physics help. three problems.



Deathslayer7
2009-11-22, 08:11 PM
the hard one first:

Physicists first attempted to understand the hydrogen atom by applying the laws of classical physics. Consider an electron of mass m and charge -e in a circular orbit of radius r around a proton of charge +e.

Use Newtonian physics to show that the total energy of the atom is

-e^2/(8pi*r*epsilon_knot) read as negative e squared divided by eight pi times epsilon subscript knot times the radius.


problem # 2:

Balmer discovered the famous formula that bears his name by inspection and trial-and-error. See if you can discover the formula for each of the following series of wavelengths. Each formula involved an integer n, but, as in the Balmer formula, n may not start with 1.

a)125.00, 31.25, 13.89, 7.81, and 5.00 nm.

b)375, 900, 1575, 2400, 3375, and 4500 nm.



problem # 3:

For what wavelength of light does a 80.0 mW laser deliver 2.40 x 10^17 photons per s?


While answers are nice and all, I need to know how to do these problems cause i have a test tomorrow. I'm not sure where to start on each of the problems, so equations help. Thanks for any and all help, even if its only with 1, that's one closer.

Generic Archer
2009-11-22, 09:34 PM
I'm too lazy to actually do the working, but

1) attraction of two particles = whatever it is...
v when moving in a circle can be worked through from the tangential force (found above)
KE of outside particle = total KE of system (assume system stationary around centre proton)


2) I have no idea what a Balmer formula is...

3) E = hc/lamda
80e^-3 = hc /Lamda
lamda = hc/80e^-3


Dane

Deathslayer7
2009-11-22, 10:04 PM
for 1. it's that equation which i cant remember.

for 2. Balmer series is something along the lines 91.18/(1/m^2-1/n^2).

for 3. i got that. (6.63x10^(-34))*(3.0*10^8)/80 x 10^(-3) =

2.49 x 10^(-25) meters = 2.49 x 10^(-15) nm (gave me a wrong answer).

The Extinguisher
2009-11-22, 10:18 PM
Here's what I got for three.

80.0 mW = 0.08 W
=0.08 J/s

So the energy of 2.40e17 photons is 0.08 J.
Energy of one photon is 3.33e-19
Then use E = hc/λ
λ = hc/E
= (6.63e-34)(3.00e8)/(3.33e-19)
= 5.97e-6 m
= 597 nm (orange light)


And for one,

F(electrostatic) = (k*q1*q2)/(r^2)
F(centripetal) = (m*v^2)/(r)

Deathslayer7
2009-11-22, 10:47 PM
where k= what exactly?

edit: nvm its 1/4piepsilon_knot