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CP Consider a model of a hydrogen atom in which an electron is in a circular orbit of radius \(r=5.29 \times 10^{-11} \mathrm{m}\) around a stationary proton. What is the speed of the electron in its orbit?

Short Answer

Expert verified
The speed of the electron is approximately \(2.18 \times 10^6 \, \text{m/s}\).

Step by step solution

01

Understand the problem

We need to find the speed of an electron revolving around a proton in a hydrogen atom. The given radius of the electron's orbit is \(5.29 \times 10^{-11} \text{m}\). This problem involves using physics concepts related to centripetal force and electrostatic force.
02

Identify the relevant formulas

In a hydrogen atom, the electrostatic force provides the necessary centripetal force for the electron to move in a circular orbit. The formulas involved are:1. Electrostatic force, given by Coulomb's Law: \( F_{e} = \frac{k \cdot e^2}{r^2} \)2. Centripetal force required for circular motion: \(F_{c} = \frac{m_e v^2}{r}\)where \(k\) is Coulomb’s constant \(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\), \(e\) is the elementary charge \(1.6 \times 10^{-19} \, \text{C}\), \(m_e\) is the mass of the electron \(9.11 \times 10^{-31} \, \text{kg}\), and \(v\) is the speed of the electron.
03

Equate the forces for the electron

Since the electrostatic force provides the necessary centripetal force, we equate the two:\[ \frac{k \cdot e^2}{r^2} = \frac{m_e v^2}{r} \]
04

Solve the equation for velocity

Rearrange the equation to solve for the speed \(v\) of the electron:Multiply both sides by \(r\):\[ \frac{k \cdot e^2}{r} = m_e v^2 \]Solve for \(v^2\):\[ v^2 = \frac{k \cdot e^2}{m_e \, r} \]Take the square root to find \(v\):\[ v = \sqrt{\frac{k \cdot e^2}{m_e \, r}} \]
05

Plug in the values and calculate

Substitute \(k = 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\), \(e = 1.6 \times 10^{-19} \, \text{C}\), \(m_e = 9.11 \times 10^{-31} \, \text{kg}\), and \(r = 5.29 \times 10^{-11} \, \text{m}\) into the equation:\[ v = \sqrt{\frac{(8.99 \times 10^9) \times (1.6 \times 10^{-19})^2}{9.11 \times 10^{-31} \times 5.29 \times 10^{-11}}} \]Calculate the value to find:\[ v \approx 2.18 \times 10^{6} \, \text{m/s} \]
06

Verify the units

Check that all units are consistent:- The numerator has units of \((\text{Nm}^2/\text{C}^2)(\text{C}^2) = \text{Nm}^2\).- The denominator has units of \((\text{kg})(\text{m}) = \text{kg} \cdot \text{m}\).- After simplification for speed, we have \(\text{m}^2/\text{s}^2\), which gives speed units of \(\text{m/s}\) after taking the square root. The calculation is consistent.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electrostatic Force
In a hydrogen atom, the electrostatic force is the attractive force between the negatively charged electron and the positively charged proton. This force acts like a binder, keeping the electron in place as it orbits around the proton. The strength of this force is determined by Coulomb's Law and is given by the formula:\[ F_{e} = \frac{k \cdot e^2}{r^2} \]Here:
  • \( k \) is Coulomb's constant and has a value of \(8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2\).
  • \( e \) is the charge of the electron or proton, amounting to \(1.6 \times 10^{-19} \, \text{C}\).
  • \( r \) is the distance between the electron and proton, which is the given radius of the electron’s orbit.
The electrostatic force is essential in holding together atoms and is much stronger than the gravitational force at the atomic scale. This force also acts as the centripetal force that keeps the electron in a stable orbit around the proton.
Centripetal Force
Centripetal force is the force that keeps an object moving in a circular path, directed towards the center of the circle. In the context of a hydrogen atom, this force is what enables the electron to orbit the nucleus, which in this case, is the stationary proton.The expression for centripetal force in circular motion is:\[ F_{c} = \frac{m_e v^2}{r} \]Where:
  • \( m_e \) is the mass of the electron, approximately \(9.11 \times 10^{-31} \, \text{kg}\).
  • \( v \) is the velocity of the electron.
  • \( r \) is the radius of the circular path that the electron follows.
The centripetal force necessary for the electron’s motion is exactly provided by the electrostatic force as per Step 3 of the solution. This relationship highlights the close interplay between the motion and forces acting on subatomic particles. Ensuring that these forces are in a delicate balance allows the electron to maintain a stable orbit, analogous to the way a planet orbits a star.
Electron Velocity
Electron velocity is a measure of how fast the electron orbits the proton in the hydrogen atom. By treating the electrostatic force as the source of the necessary centripetal force, we can derive the velocity equation from equating those two forces:\[ \frac{k \cdot e^2}{r^2} = \frac{m_e v^2}{r} \]Through rearranging and solving for velocity \( v \), we get:\[ v = \sqrt{\frac{k \cdot e^2}{m_e \, r}} \]This equation shows how the velocity depends on constants such as the charge \( e \), mass \( m_e \), and Coulomb’s constant \( k \). By substituting the known values into the formula, the velocity of the electron is calculated to be approximately \(2.18 \times 10^6 \, \text{m/s}\).Understanding electron velocity is crucial as it affects energy levels and the radii of electron orbits. High velocities imply more energetic states, which are reflected in the quantum mechanics that underpin atomic physics. The classical interpretation taught here provides a stepping stone towards the more comprehensive quantum mechanical descriptions.

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Most popular questions from this chapter

A particle has charge \(-3.00 \mathrm{nC}\) . (a) Find the magnitude and direction of the electric field due to this particle at a point 0.250 m directly above it. (b) At what distance from this particle does its electric field have a magnitude of 12.0 \(\mathrm{N} / \mathrm{C} ?\)

A point charge is at the origin. With this point charge as the source point, what is the unit vector \(r\) in the direction of (a) the field point at \(x=0, \quad y=-1.35 \mathrm{m}\) ; (b) the field point at \(x=12.0 \mathrm{cm}, y=12.0 \mathrm{cm} ;(\mathrm{c})\) the field point at \(x=-1.10 \mathrm{m},\) \(y=2.60 \mathrm{m} ?\) Express your results in terms of the unit vectors \(\hat{\boldsymbol{\imath}}\) and \(\hat{\boldsymbol{J.}}\)

CP Two identical spheres are each attached to silk threads of length \(L=0.500 \mathrm{m}\) and hung from a common point (Fig. P21.68). Each sphere has mass \(m=8.00 \mathrm{g} .\) The radius of each sphere is very small compared to the distance between the spheres, so they may be treated as point charges. One sphere is given positive charge \(q_{1},\) and the other a different positive charge \(q_{2} ;\) this causes the spheres to separate so that when the spheres are in equilibrium, each thread makes an angle \(\theta=20.0^{\circ}\) with the vertical. (a) Draw a free-body diagram for each sphere when in equilibrium, and label all the forces that act on each sphere. (b) Determine the magnitude of the electrostatic force that acts on each sphere, and determine the tension in each thread. (c) Based on the information you have been given, what can you say about the magnitudes of \(q_{1}\) and \(q_{2} ?\) Explain your answers. (d) A small wire is now connected between the spheres, allowing charge to be transferred from one sphere to the other until the two spheres have equal charges; the wire is then removed. Each thread now makes an angle of \(30.0^{\circ}\) with the vertical. Determine the original charges. (Hint: The total charge on the pair of spheres is conserved.)

Point charges \(q_{1}=-4.5 \mathrm{nC}\) and \(q_{2}=+4.5 \mathrm{nC}\) are separated by 3.1 mm, forming an electric dipole. (a) Find the electric dipole moment (magnitude and direction). (b) The charges are in a uniform electric field whose direction makes an angle of \(36.9^{\circ}\) with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude \(7.2 \times 10^{-9} \mathrm{N} \cdot \mathrm{m} ?\)

\(A+2.00-n C \quad\) point charge is at the origin, and a second \(-5.00\) -n \(\mathrm{C}\) point charge is on the \(x\) -axis at \(x=0.800 \mathrm{m}\) . (a) Find the electric field (magnitude and direction) at each of the following points on the \(x\) -axis: (i) \(x=\) \(0.200 \mathrm{m} ;\) (ii) \(x=1.20 \mathrm{m} ;\) (iii) \(x=-0.200 \mathrm{m} .\) (b) Find the net electric force that the two charges would exert on an electron placed at each point in part (a).

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