Question

What two pieces of evidence allowed the first calculation of m_e, the mass of the electron?

(a) The ratios $$\frac{{\mathbf{q}}_{\mathbf{e}}}{{\mathbf{m}}_{\mathbf{e}}}$$and $$\frac{{\mathbf{q}}_{\mathbf{p}}}{{\mathbf{m}}_{\mathbf{p}}}$$.

(b) The values of q_e and E_B.

(c) The ratio $$\frac{{\mathbf{q}}_{\mathbf{e}}}{{\mathbf{m}}_{\mathbf{e}}}$$and q_e.

Justify your response.

Short answer

Option (c) once the ratio $$\frac{q_e}{m_e}$$ and the magnitude of the electric charge q_e was known, it was possible to calculate the mass of an electron m_e .

Step-by-step solution

Concept Introduction

Electrons are subatomic particles having an elementary charge of- 1. An electron and a proton both have the same amount of charge (but has an opposite sign).

Determine the formulas

Consider the formula for the mas of the electron as:

$${\mathbf{m}}_{\mathbf{e}}\mathbf{=}\frac{{\mathbf{q}}_{\mathbf{e}}}{\left(\frac{{\mathbf{q}}_{\mathbf{e}}}{{\mathbf{m}}_{\mathbf{e}}}\right)}$$

Determine the mass of electron

The ratio$$\frac{q_e}{m_e}$$was calculated using the amount of acceleration beam deflection caused by the application of a magnetic field to cathode-ray tubes.

In the future, Millikan conducted the oil drop experiment that produced the value of q_e . It was feasible to determine the mass of an electronm_e from these two values.

For example -

The charge of electron is - $$q_e=-1.60\times {10}^{-19}C$$.

The mass of the electron is calculated as:

$$\begin{array}{c}{m}_e=\frac{-1.60\times {10}^{-18}\mathrm{C}}{-1.76\times {10}^{11}\frac{\mathrm{C}}{\mathrm{k}\mathrm{g}}}\\ =9.11\times {10}^{-31}\mathrm{k}\mathrm{g}\end{array}$$

Therefore, the ratio $$\frac{q_e}{m_e}$$ and q_e is used to calculate mass of electron.