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Chapter 1: Problem 21

The neutral hydrogen atom in its normal state behaves in some respects like an electric charge distribution which consists of a point charge of magnitude \(e\) surrounded by a distribution of negative charge whose density is given by \(-\rho(r)=C e^{-2 r / a_{0}} .\) Here \(a_{0}\) is the Bohr radius, \(0.53 \times 10^{-8} \mathrm{~cm}\), and \(C\) is a constant with the value required to make the total amount of negative charge exactly \(e .\) What is the net electric charge inside a sphere of radius \(a_{0}\) ? What is the electric field strength at this distance from the nucleus?

Short Answer

Expert verified
The net charge is \(ee^{-2}\) and the electric field at distance \(a_0\) is \(\frac{e e^{-2}}{4\pi \varepsilon_0 a_0^2}\).

Step by step solution

01

Calculate Total Negative Charge

The negative charge density is given by \(-\rho(r) = C e^{-2 r / a_0}\). To find the total negative charge, integrate the density over all space. The negative charge must sum to \(-e\). The constant \(C\) is chosen such that the integral of the density over all space equals \(-e\) in order to satisfy the condition of charge neutrality.
02

Determine the Constant C

We need to solve for C by integrating the density over all space and setting it equal to -e: \ \[\int_{0}^{\infty} 4\pi r^2 \left(C e^{-2r/a_0}\right) dr = -e.\]This integral evaluates to \[C \frac{4\pi a_0^3}{8} = e\]which simplifies to \[C \times \frac{\pi a_0^3}{2} = e.\]Solving, we find\[C = \frac{2e}{\pi a_0^3}.\]
03

Net Charge Inside Sphere of Radius a0

Integrate the negative charge density from 0 to \(a_0\) to find the negative charge within the sphere:\[Q_{neg} = \int_{0}^{a_0} 4\pi r^2 \left(C e^{-2r/a_0}\right) dr\]Substitute \(C = \frac{2e}{\pi a_0^3}\):\[Q_{neg} = \frac{2e}{a_0^3} \int_{0}^{a_0} 4\pi r^2 e^{-2r/a_0} dr.\]Solving using the exponential integral gives\[Q_{neg} = -e (1 - e^{-2}).\]The net charge is\[Q_{net} = e + Q_{neg} = e - e(1 - e^{-2}) = ee^{-2}.\]
04

Calculate Electric Field at Radius a0

Using Gauss's Law, the electric field \(E\) at distance \(r = a_0\) is given by\[E \times 4\pi a_0^2 = \frac{Q_{net}}{\varepsilon_0}.\]Solve for \(E\):\[E = \frac{e e^{-2}}{4\pi \varepsilon_0 a_0^2}.\]

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Key Concepts

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

Electric Charge Distribution
In a neutral hydrogen atom, we can imagine the electric charge distribution as two distinct parts: a positive point charge at the nucleus and a surrounding cloud of negative charge. This model helps us understand how electric forces behave within the atom. The point charge has a magnitude of exactly one elementary charge, denoted as \( e \).
The negative charge distribution is where things get a bit more complex. Its density is represented by a function, \(-\rho(r) = C e^{-2r/a_0}\). Here, \( C \) is a constant that adjusts to ensure the total negative charge equals \( -e \). The exponential part, \( e^{-2r/a_0} \), describes how the charge density decreases as the distance \( r \) from the nucleus increases.
  • The function \(-\rho(r)\) indicates the local density of negative charge at any point \( r \).
  • The total negative charge in the atom must be \(-e\) to balance the positive charge, ensuring electrical neutrality.
  • This charge distribution is crucial in calculations involving the electric fields and forces at various distances from the nucleus.
Bohr Radius
The Bohr radius, denoted as \( a_0 \), is a fundamental constant in atomic physics, approximately equal to \( 0.53 \times 10^{-8} \) cm. In the Bohr model of the hydrogen atom, it represents the average distance between the electron and the nucleus.
The Bohr radius plays a vital role in defining the scale of atomic structures and calculations involving electric charge distributions. When calculating internal electric forces and fields, the Bohr radius is the natural unit of measure due to its relevance to the size of the hydrogen atom.
  • In many calculations, such as finding the net charge within a sphere of radius \( a_0 \), this value helps determine key properties of the hydrogen atom.
  • The concept of the Bohr radius is essential for understanding the balance between attractive and repulsive forces in hydrogen atoms.
  • It allows for a simplified view of atomic orbits, making it an important artifact of early quantum mechanics models that helps bridge classical physics with quantum behavior.
Gauss's Law
In electromagnetism, Gauss's Law is a powerful tool for calculating electric fields when certain symmetries are present. It relates the distribution of electric charge to the resulting electric field. In the context of our hydrogen atom problem, Gauss's Law helps compute the electric field at a certain distance from the nucleus.
According to Gauss's Law, the electric field \( E \) at a distance \( r \) from the center of an electric charge distribution can be expressed as: \[ E imes 4\pi r^2 = \frac{Q_{net}}{\varepsilon_0} \]where \( Q_{net} \) is the net charge enclosed within a sphere of radius \( r \), and \( \varepsilon_0 \) is the permittivity of free space.
  • Gauss's Law simplifies problems with symmetrical charge distributions, such as those in spherical or cylindrical geometries.
  • By using it, we can find the electric field at the Bohr radius by knowing the net charge contained within that region.
  • This principle underpins much of electrostatics and allows for efficient calculations with manageable mathematical effort.

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Most popular questions from this chapter

Consider a spherical charge distribution which has a constant density \(\rho\) from \(r=0\) out to \(r=a\) and is zero beyond. Find the electric field for all values of \(r\), both less than and greater than \(a\). Is there a discontinuous change in the field as we pass the surface of the charge distribution at \(r=a ?\) Is there a discontinuous change at \(r=0 ?\)

A sphere of radius \(R\) has a charge \(Q\) distributed uniformly over its surface. How large a sphere contains 90 percent of the energy stored in the electrostatic field of this charge distribution?

A spherical volume of radius \(a\) is filled with charge of uniform density \(\rho .\) We want to know the potential energy \(U\) of this sphere of charge, that is, the work done in assembling it. Calculate it by building the sphere up layer by layer, making use of the fact that the field outside a spherical distribution of charge is the same as if all the charge were at the center. Express the result in terms of the total charge \(Q\) in the sphere. Ans. \(U=\Varangle\left(Q^{2} / a\right)\).

At each corner of a square is a particle with charge \(q\). Fixed at the center of the square is a point charge of opposite sign, of magnitude \(Q\). What value must \(Q\) have to make the total force on each of the four particles zero? With \(Q\) set at that value, the system, in the absence of other forces, is in equilibrium. Do you think the equilibrium is stable? Ans. \(Q=0.957 q .\)

Suppose three positively charged particles are constrained to move on a fixed circular track. If the charges were all equal, an equilibrium arrangement would obviously be a symmetrical one with the particles spaced \(120^{*}\) apart around the circle. Suppose that two of the charges are equal and the equilibrium arrangement is such that these two charges are \(90^{\circ}\) apart rather than \(120^{\circ}\). What is the relative magnitude of the third charge?
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