The smallest unit of charge is − 1.6 × 10 − 19 C, −1.6×10−19 C, which is the charge in coulombs of a single electron. Robert Millikan was able to measure the charge on small droplets of oil by suspending them between a pair of electrically charged plates. Which of the values are possible charges of those oil droplets?

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The question is incomplete, here is the complete question:

The smallest unit of charge is - 1.6 × 10⁻¹⁹ C, which is the charge in coulombs of a single electron.

Robert Millikan was able to measure the charge on small droplets of oil by suspending them between a pair of electrically charged plates. Which of the following are possible charges of those oil droplets?

A. - 1.2 × 10⁻¹⁹ C

B. - 3.2 × 10⁻¹⁹ C

C. - 4.8 × 10⁻¹⁹ C

D. - 5.6 × 10⁻¹⁹ C

E. - 8.0 × 10⁻¹⁹ C

F. - 9.4 × 10⁻¹⁹ C

Answer: The possible charges of the oil droplets are [tex]-3.2\times 10^{-19}C[/tex] , [tex]-4.8\times 10^{-19}C[/tex]  and [tex]-8.0\times 10^{-19}C[/tex]

Explanation:

Millikan’s oil drop experiment is used to measure the charge of an electron. Before this experiment, the subatomic particles were not accepted.

He found that all the oil drops had charges that were the multiples of [tex]-1.6\times 10^{-19}C[/tex]. This value is the charge on 1 electron

Number of electrons excess electrons is calculated by using the formula:

[tex]\text{Excess electrons}=\frac{\text{Charge on millikan's oil drop}}{\text{Charge on 1 electron}}[/tex]

From the given options:

  • Option A: [tex]-1.2\times 10^{-19}C[/tex]

[tex]\text{Excess electrons}=\frac{-1.2\times 10^{-19}}{-1.6\times 10^{-19}}=0.75[/tex]

As, the excess electron is not a whole number. So, this charge cannot be taken as the charge on an oil drop.

  • Option B: [tex]-3.2\times 10^{-19}C[/tex]

[tex]\text{Excess electrons}=\frac{-3.2\times 10^{-19}}{-1.6\times 10^{-19}}=2[/tex]

As, the excess electron is a whole number. So, this charge can be taken as the charge on an oil drop.

  • Option C: [tex]-4.8\times 10^{-19}C[/tex]

[tex]\text{Excess electrons}=\frac{-4.8\times 10^{-19}}{-1.6\times 10^{-19}}=3[/tex]

As, the excess electron is a whole number. So, this charge can be taken as the charge on an oil drop.

  • Option D: [tex]-5.6\times 10^{-19}C[/tex]

[tex]\text{Excess electrons}=\frac{-5.6\times 10^{-19}}{-1.6\times 10^{-19}}=3.5[/tex]

As, the excess electron is not a whole number. So, this charge cannot be taken as the charge on an oil drop.

  • Option E: [tex]-8.0\times 10^{-19}C[/tex]

[tex]\text{Excess electrons}=\frac{-8.0\times 10^{-19}}{-1.6\times 10^{-19}}=5[/tex]

As, the excess electron is a whole number. So, this charge can be taken as the charge on an oil drop.

  • Option F: [tex]-9.4\times 10^{-19}C[/tex]

[tex]\text{Excess electrons}=\frac{-9.4\times 10^{-19}}{-1.6\times 10^{-19}}=5.875[/tex]

As, the excess electron is not a whole number. So, this charge cannot be taken as the charge on an oil drop.

Hence, the possible charges of the oil droplets are [tex]-3.2\times 10^{-19}C[/tex] , [tex]-4.8\times 10^{-19}C[/tex]  and [tex]-8.0\times 10^{-19}C[/tex]