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Chapter 30: Q4CQ (page 1110)

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

(a) The ratios \[\frac{{{{\bf{q}}_{\bf{e}}}}}{{{{\bf{m}}_{\bf{e}}}}}\]and \[\frac{{{{\bf{q}}_{\bf{p}}}}}{{{{\bf{m}}_{\bf{p}}}}}\].

(b) The values of qe and EB.

(c) The ratio \[\frac{{{{\bf{q}}_{\bf{e}}}}}{{{{\bf{m}}_{\bf{e}}}}}\]and qe.

Justify your response.

Short Answer

Expert verified

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

Step by step solution

01

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).

02

Determine the formulas

Consider the formula for the mas of the electron as:

\[{{\bf{m}}_{\bf{e}}}{\bf{ = }}\frac{{{{\bf{q}}_{\bf{e}}}}}{{\left( {\frac{{{{\bf{q}}_{\bf{e}}}}}{{{{\bf{m}}_{\bf{e}}}}}} \right)}}\]

03

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 qe . It was feasible to determine the mass of an electronme 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:

\[\begin{array}{c}{m_e} = \frac{{ - 1.60 \times {{10}^{ - 18}}\;{\rm{C}}}}{{ - 1.76 \times {{10}^{11}}\;\frac{{\rm{C}}}{{{\rm{kg}}}}}}\\ = 9.11 \times {10^{ - 31}}\;\;{\rm{kg}}\end{array}\]

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

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Most popular questions from this chapter

Crystal lattices can be examined with x rays but not UV. Why?

Particles called muons exist in cosmic rays and can be created in particle accelerators. Muons are very similar to electrons, having the same charge and spin, but they have a mass 207 times greater. When muons are captured by an atom, they orbit just like an electron but with a smaller radius, since the mass in\({{\bf{a}}_{\bf{B}}}{\bf{ = }}\frac{{{{\bf{h}}^{\bf{2}}}}}{{{\bf{4}}{{\bf{\pi }}^{\bf{2}}}{{\bf{m}}_{\bf{e}}}{\bf{kq}}_{\bf{e}}^{\bf{2}}}}{\bf{ = 0}}{\bf{.529 \times 1}}{{\bf{0}}^{{\bf{ - 10}}}}{\bf{\;m is 207}}{{\bf{m}}_{\bf{e}}}\)

(a) Calculate the radius of then = 1 orbit for a muon in a uranium ion\(\left (Z = 92).

(b) Compare this with the7.5 - fmradius of a uranium nucleus. Note that since the muon orbits inside the electron, it falls into a hydrogen-like orbit. Since your answer is less than the radius of the nucleus, you can see that the photons emitted as the muon falls into its lowest orbit can give information about the nucleus.

Integrated Concepts

A galaxy moving away from the earth has a speed of 0.0100c. What wavelength do we observe for an ni = 7 to nf = 2 transition for hydrogen in that galaxy?

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(a) Verify that the energy of the photon emitted when neon goes from its metastable state to the one immediately below is equal to 1.96 eV.

(b) Show that the wavelength of this radiation is 633 nm.

(c) What wavelength is emitted when the neon makes a direct transition to its ground state?

If a hydrogen atom has its electron in the n = 4 state, how much energy in eV is needed to ionize it?
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