Answer to Question #105753 in Molecular Physics | Thermodynamics for Sophie

Question

Millikan's oil drop experiment the mass of the oil drops is obtained by observing the terminals Vt of a freely falling drop in the absence of an electric field. under the circumstances the effective weight equals the viscous Force given by stroke's law F= 6ie ΠrVr where Π is 1.8 * 10 - 5 kg/m.s is the viscosity of air and r is 0.65 cm is the of the radius of the drop . Also the actual with Mg =⁴÷² pie r³ density gram. must be corrected for the buoyant force of the air this is done by replacing density with density -densityA .density is 9.20 x 10² kg/m³ the density of the oil and density is 1.29 kg/m³ density of air with this premier a show that charge drop is given by

Answer 1

As per the given question,

$F=6\pi \eta rV_t$

$\eta= 1.8\times 10^{ - 5} kg/m.s$

radius (r)=0.65cm

Density of oil drop $\rho_{oil}=9.20 \times 10^2 kg/m^3$

Density of the air$\rho_{air}=1.29 \times 10^{2}kg/m^3$

We know that,

downward force =Upward force

Let the mass of the drop is M and g is the gravitational acceleration, $F_{th}$ is the thrust force on the drop, let V is the volume of the drop is V,

So, $Mg-F_{th}=6\pi \eta rV_t$

$\Rightarrow V_t=\dfrac{Mg-\rho_{air}\times V}{6\pi \eta r}$

$\Rightarrow V_t=\dfrac{V(\rho_{oil}-\rho_{air})}{6\pi \eta r}$

Now, substituting the values in the above,

$V_t=\dfrac{(\rho_{oil}-\rho_{air}) \times \dfrac{4}{3}\times \pi r^3}{6\pi \eta r}$

$\Rightarrow V_t=\dfrac{2(\rho_{oil}-\rho_{air}) \pi r^2}{9\pi\times \eta}=\dfrac{(\rho_{oil}-\rho_{air}) r^2}{9\eta}$

$\Rightarrow V_t=\dfrac{2\times(9.20-1.29)\times 10^{2}\times 0.65\times 10^{-2}}{6\times 1.8\times 10^{-5}}$ m/sec

$V_t=\dfrac{2\times 7.91\times 0.65}{9\times 10^{-5}}=\dfrac{10.283\times 10^{5}}{9}m/sec$

$\Rightarrow V_t=1.14\times10^{5} m/sec$