In hydrogen’s characteristic spectra, each series - the Lyman, the Balmer, and so on – has a “series limit,” where the wavelengths at one end of the series tend to bunch up, approaching a single limiting value. Is it at the short-wavelength or the long-wavelength end of the series that the series limit occurs, and what is it about hydrogen’s allowed energies that leads to this phenomenon? Does the infinite well have series limits?

Short Answer

Expert verified

The short-frequency end. No, the boundless well does not have series limits.

Step by step solution

01

Define the energy of the hydrogen atom and the energy in a 3-D potential.

The allowed energy of the hydrogen atom is,

En=-me42(4πε0)2h2n

The allowed energy in a 3D potential well is,

Enx,ny,nz=(nx2Lx2+ny2Ly2+nz2Lz2)π2h22m

Here n is the integer, m78C8 is the mass of the electron, e is the charge of the electron, nx , ny, nz, are integers and ε0is the permittivity of free space.

02

Calculation of the energy of the hydrogen atom

Taking a gander at the recipe for the permitted energies of the hydrogen particle from the condition,

En=-me424πε02h2n

Thus, the ground level has the best bad energy worth, and afterward, the worth declines by a component of . The distinction between the energy levels declines too, as gets increasingly elevated.

As approaches vastness here, the energy of the electron becomes zero, which implies that the electron is free and not bound to the particle any longer.

03

Calculation of the energy in a 3-D potential well

Taking a gander at the recipe for the permitted energies in a 3D-boundless likely well from the condition,

Enx,ny,nz=(nx2Lx2+ny2Ly2+nz2Lz2)π2h22m

For a cubic well, and taking nx2+ny2+nz2=n2this becomes,

En=n2π2h22mL

Thus, the ground level has the most reduced energy worth, and afterward, the worth increments by a variable of . The distinction between the energy levels also increases as n ever more elevated.

Here, the molecule cannot have zero energy. The most minimal energy is the ground state energy or the zero-point energy.

04

The occurrence of the series limit and its phenomenon

From the above figure, in the hydrogen molecule, the levels bundle up (draw nearer to one another) at the top. Thusly, as far as possible will relate to the most noteworthy energy progress in a specific series (ionization energy). Since energy is conversely relative to the frequency as indicated by E=hcλ.

Consequently, as far as possible is at the short-wavelength end of the series. For the limitless well, there is no "bundling up" of the levels, as they get further separated from one another, not nearer as n increments. In like manner, the endless well doesn't have a series limit like that of the hydrogen iota.

Hence, the short-frequency end. No, the boundless well does not have series limits.

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Most popular questions from this chapter

Mathematically equation (7-22) is the same differential equation as we had for a particle in a box-the function and its second derivative are proportional. But(ϕ)for m1= 0is a constant and is allowed, whereas such a constant wave function is not allowed for a particle in a box. What physics accounts for this difference?

A particular vibrating diatomic molecule may be treated as a simple harmonic oscillator. Show that a transition from that n=2state directly to n=0ground state cannot occur by electric dipole radiation.

A hydrogen atom electron is in a 2p state. If no experiment has been done to establish a z-component of angular momentum, the atom is equally likely to be found with any allowed value of LZ. Show that if the probability densities for these different possible states are added (with equal weighting), the result is independent of both ϕandθ

A hydrogen atom in an n = 2 state absorbs a photon,

  1. What should be the photon wavelength to cause the electron to jump to an n = 4 state?
  2. What wavelength photons might be emitted by the atom following this absorption?

A wave function with a non-infinite wavelength-however approximate it might be- has nonzero momentum and thus nonzero kinetic energy. Even a single "bump" has kinetic energy. In either case, we can say that the function has kinetic energy because it has curvature- a second derivative. Indeed, the kinetic energy operator in any coordinate system involves a second derivative. The only function without kinetic energy would be a straight line. As a special case, this includes a constant, which may be thought of as a function with an infinite wavelength. By looking at the curvature in the appropriate dimension(s). answer the following: For a givenn,isthe kinetic energy solely

(a) radial in the state of lowest l- that is, l=0; and

(b) rotational in the state of highest l-that is, l=n-1?

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