Q.2.32 P - Chemistry: Molecular Nature Of Matter And Change

Question

Thomson was able to determine the mass/charge ratio of the electron but not it's mass. How did Millikan’s experiment allow the determination of the electron’s mass?

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 $\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{\times }{\mathbf{10}}^{\mathbf{19}}\mathit{C}$. Mass of electron is calculated by using mass charge ratio given by Thomson that comes out to be
 $\mathbf{9}\mathbf{.}\mathbf{109}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{31}}\mathit{K}\mathit{g}$

Step 1: Mass charge ratio of the electron:

According toThomson mass /charge ratio of the electron is $\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{686}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{12}}$

Step 2:Millikan’s oil drop experiment:

Millikan performed an oil drop experiment in which he measured the mass of a droplet from its rate of fall. that he made the fall slowly, rise, or suspended by turning on the magnetic field and varying its strength, by 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 $\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{\times }{\mathbf{10}}^{\mathbf{19}}\mathit{C}$where stands for coulomb, the S.I unit of charge.

Step3:Calculation of mass of electron:

$$\begin{gathered} \mathit{M}\mathit{a}\mathit{s}\mathit{s}\mathbf{ }\mathit{o}\mathit{f}\mathbf{ }\mathit{e}\mathit{l}\mathit{e}\mathit{c}\mathit{t}\mathit{r}\mathit{o}\mathit{n}\mathbf{=}\frac{\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}}{\mathbf{c}\mathbf{h}\mathbf{a}\mathbf{r}\mathbf{g}\mathbf{e}}\mathbf{\times }\mathit{c}\mathit{h}\mathit{a}\mathit{r}\mathit{g}\mathit{e} \\ \mathbf{=}\mathbf{(}\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{686}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\frac{\mathbf{K}\mathbf{g}}{\mathbf{C}}\mathbf{)}\mathbf{(}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{602}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{C}\mathbf{)} \\ \mathbf{=}\mathbf{9}\mathbf{.}\mathbf{109}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{31}}\mathit{K}\mathit{g}\mathbf{ }\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{109}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{28}}\mathit{g} \end{gathered}$$

Result:

Hence the mass of the electron calculated by Millikan after his experiment is

$\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{109}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{31}\mathbf{ }}\mathit{K}\mathit{g}\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{9}\mathbf{.}\mathbf{109}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{28}}\mathit{g}$