Q. 2.33 P - Chemistry: Molecular Nature Of Matter And Change

Question

The following charges on individual oil droplets were obtained during an experiment similar to Millikan's. Determine a charge for the electron( in coulombs), and explain your answer:
$\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{\times }{\mathbf{10}}^{\mathbf{19}}\mathit{C}$ ; $\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{806}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathit{C}$; $\mathbf{-}\mathbf{8}\mathbf{.}\mathbf{010}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathit{C}$; $\mathbf{-}\mathbf{14}\mathbf{.}\mathbf{42}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathit{C}$.

Step-by-Step Solution

Answer

Millikan performed an oil drop experiment and observed charges of oil droplets were whole number multiple of a minimum charge that is $\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{\times }{\mathbf{10}}^{\mathbf{19}}\mathit{C}$. The charges obtained during an experiment similar to Millikan’s experiment are multiple of $\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{\times }{\mathbf{10}}^{\mathbf{19}}\mathit{C}$

Step 1: Chargetheelectron:

According to the Millikan oil drop’s experiment charge of the electron is $\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathit{C}$

Step 2: Millikan’s oil drop experiment:

In his oil drop experiment, Millikan measured the mass of a droplet from its rate of fall, that he made the fall slowly, rise, or be suspended by turning on the magnetic field and varying its strength, by doing this, he calculated the total charge on the droplet. 

He found that various charges of droplets are always whole number multiple of a minimum charge. The value he calculated of the electric charge is $\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathit{C}$ where stands for coulomb, the S.I unit of charge.

Step 3: Calculation of mass of electron:

The charges obtained during an experiment similar to Millikan’s experiment are multiple $\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{\times }{\mathbf{10}}^{\mathbf{19}}\mathit{C}$.

$$\begin{gathered} \mathbf{-}\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathit{C}\mathbf{\times }\mathbf{2}=-3.204\times {10}^{-19}C \\ \mathbf{-}\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{\times }\mathbf{3}= -4.806\times {10}^{-19}C \\ \mathbf{-}\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{\times }\mathbf{5} = -8.010\times {10}^{-19}C \\ \mathbf{-}\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{\times }\mathbf{9} = -14.42\times {10}^{-19}C \end{gathered}$$

Result:

 Hence, the charges obtained during an experiment similar to Millikan’s experiment are multiple 

$\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathit{C}$i.e. $\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{204}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathit{C}$, $\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{806}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathit{C}$,$\mathbf{-}\mathbf{8}\mathbf{.}\mathbf{010}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathit{C}$,$\mathbf{-}\mathbf{14}\mathbf{.}\mathbf{42}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathit{C}$.