Physics307L F08:People/Gibson/Millikan Oil Drop Summary

From OpenWetWare

Jump to: navigation, search

Purpose

The purpose of this experiment was to overall determine the charge of an electron based upon several rise and fall times of oil drops. To further explain this, we introduced oil drops into the Millikan device then subjected them to alpha particles. After allowing some small margin of time to pass for the oil drops to ball in between the viewable region (in between the two charged plates) we then proceeded to introduce a charge on the drop to either make it rise or fall. Grid lines in the viewing scope allowed us to measure a distance more accurately as we used the stop watch to calculate these times.

Equations

  • These are equations relating the constants and other important parameters, that allow us to calculate the charge of each drop.

Where:

α is the radius of the drop in m,
b is a constant (8.20x10^-3 Pa*m),
p is the pressure in pascals
ρ is the density of oil in kg/m^3
η is the viscosity of air in poise (Ns/m^2)
g is the acceleration of gravity in m/s^2
Vf is the velocity of fall in m/s


\alpha=\sqrt{\frac{b^{2}}{{4p}^{2}}+\frac{9nV_f}{2g\rho}}-\frac{b}{2p}
m=\frac{4}{3}\pi\left(\sqrt{\frac{b^{2}}{{4p}^{2}}+\frac{9nV_f}{2g\rho}}-\frac{b}{2p}\right)^{3}\rho
E=\frac{V}{300d}
Q=mg\frac{\left( V_f+V_r\right)}{EV_f}
Q=\frac{4}{3}\pi\rho g\left(\sqrt{\frac{b^{2}}{{4p}^{2}}+\frac{9nV_f}{2g\rho}}-\frac{b}{2p}\right)^{3}\frac{\left( V_f+V_r\right)}{EV_f}


Results

SJK 15:35, 18 November 2007 (CST)
15:35, 18 November 2007 (CST)need a link to your primary lab notebook!
15:35, 18 November 2007 (CST)
need a link to your primary lab notebook!
  • for this experiment we found that the elementary charge(i.e the charge of a single electron) to be
SJK 16:12, 18 November 2007 (CST)
16:12, 18 November 2007 (CST)looking at your primary notebook and this summary, I have no clue how you obtain this value!
16:12, 18 November 2007 (CST)
looking at your primary notebook and this summary, I have no clue how you obtain this value!
       e=1.653x10^-19 C

and looking at the relative error with the accepted value of e, we have an error of Relative Error=

       1.60x10^-19C - 1.60x10^-19C / 1.60x10^-19=0.032 = 3.2%

From these results we're more than happy with this amount of error.

Personal tools