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Chapter 16: Problem 79

The Bohr model of the hydrogen atom proposes that the electron orbits around the proton in a circle of radius \(5.3 \times 10^{-11} \mathrm{m} .\) The electric force is responsible for the radial acceleration of the electron. What is the speed of the electron in this model?

Short Answer

Expert verified
Answer: The speed of the electron in the hydrogen atom using the Bohr model is approximately \(2.18 \times 10^6 \;\text{m/s}\).

Step by step solution

01

Write down the given information

We are given the radius of the electron's orbit in the hydrogen atom: \(r = 5.3 \times 10^{-11} \mathrm{m}\).
02

Equate the centripetal force and electric force

The centripetal force on the electron is provided by the electric force between the electron and proton. Thus, we can equate the two forces: $$m \times a_c = \frac{k \times e^2}{r^2}$$, where \(a_c = \frac{v^2}{r}\) is the centripetal acceleration, and \(v\) is the speed we are looking for.
03

Substitute the centripetal acceleration

Replace \(a_c\) with \(\frac{v^2}{r}\) in the above equation: $$m \times \frac{v^2}{r} = \frac{k \times e^2}{r^2}$$.
04

Rearrange the equation to solve for \(v\)

We need to isolate \(v\) so we can find its value: $$v^2 = \frac{k \times e^2 \times r}{m}$$.
05

Substitute the known values and constants

Apply the known values and constants to the equation: $$v^2 = \frac{(8.99\times10^9\;\mathrm{N\cdot m^2/C^2})\times (1.6\times10^{-19}\;\mathrm{C})^2 \times (5.3\times10^{-11}\;\mathrm{m})}{9.11\times10^{-31}\;\mathrm{kg}}$$.
06

Calculate the value of \(v\)

Perform the calculation to find the value of \(v\): $$v=\sqrt{\frac{(8.99\times10^9\;\mathrm{N\cdot m^2/C^2})\times (1.6\times10^{-19}\;\mathrm{C})^2 \times (5.3\times10^{-11}\;\mathrm{m})}{9.11\times10^{-31}\;\mathrm{kg}}}$$, which yields \(v \approx 2.18 \times 10^6 \;\text{m/s}\). Solution: The speed of the electron in the hydrogen atom using the Bohr model is approximately \(2.18 \times 10^6 \;\text{m/s}\).

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Most popular questions from this chapter

A flat conducting sheet of area \(A\) has a charge \(q\) on each surface. (a) What is the electric field inside the sheet? (b) Use Gauss's law to show that the electric field just outside the sheet is \(E=q /\left(\epsilon_{0} A\right)=\sigma / \epsilon_{0} .\) (c) Does this contradict the result of Problem \(69 ?\) Compare the field line diagrams for the two situations.
Two otherwise identical conducting spheres carry charges of $+5.0 \mu \mathrm{C}\( and \)-1.0 \mu \mathrm{C} .$ They are initially a large distance \(L\) apart. The spheres are brought together, touched together, and then returned to their original separation \(L\). What is the ratio of the magnitude of the force on either sphere after they are touched to that before they were touched?
The electric field across a cellular membrane is $1.0 \times 10^{7} \mathrm{N} / \mathrm{C}$ directed into the cell. (a) If a pore opens, which way do sodium ions (Na") flow- into the cell or out of the cell? (b) What is the magnitude of the electric force on the sodium ion? The charge on the sodium ion is \(+e\)
A hollow conducting sphere of radius \(R\) carries a negative charge \(-q .\) (a) Write expressions for the electric field \(\overrightarrow{\mathbf{E}}\) inside \((rR)\) the sphere. Also indicate the direction of the field. (b) Sketch a graph of the field strength as a function of \(r\). [Hint: See Conceptual Example \(16.8 .]\)
In fair weather, over flat ground, there is a downward electric field of about \(150 \mathrm{N} / \mathrm{C} .\) (a) Assume that the Earth is a conducting sphere with charge on its surface. If the electric field just outside is $150 \mathrm{N} / \mathrm{C}$ pointing radially inward, calculate the total charge on the Earth and the charge per unit area. (b) At an altitude of $250 \mathrm{m}\( above Earth's surface, the field is only \)120 \mathrm{N} / \mathrm{C}$ Calculate the charge density (charge per unit volume) of the air (assumed constant). [Hint: See Conceptual Example \(16.8 .]\)
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