/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 For an ion with nuclear charge \... [FREE SOLUTION] | 91Ó°ÊÓ

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For an ion with nuclear charge \(Z\) and a single electron, the electric potential energy is \(-\mathrm{Ze}^{2} / 4 \pi \epsilon_{0} r\) and the expression for the energies of the states and for the normalized wave functions are obtained from those for hydrogen by replacing \(e^{2}\) by \(\mathrm{Ze}^{2}\) . Consider the \(\mathrm{N}^{6+}\) ion, with seven protons and one electron. (a) What is the wavelength of the photon emitted when the \(\mathrm{N}^{6+}\) ion makes a transition from the \(n=2\) state to the \(n=1\) ground state?

Short Answer

Expert verified
The wavelength of the photon emitted is approximately 3.79 nm.

Step by step solution

01

Determine Nuclear Charge

For the nitrogen ion \(\mathrm{N}^{6+}\), there are 7 protons. Thus, the nuclear charge \(Z = 7\).
02

Apply Energy Formula for Hydrogen-like Ion

The energy levels for a hydrogen-like ion are given by \[ E_n = - \frac{Z^2 e^4 m}{8 \varepsilon_0^2 h^2 n^2} \]where \(m\) is the electron mass, \(\varepsilon_0\) is the vacuum permittivity, and \(n\) is the principal quantum number. For nitrogen \(Z = 7\).
03

Calculate Energy for\(n=2\) and \(n=1\)

Plug \(n=2\) and \(n=1\) into the energy formula:\[ E_2 = - \frac{7^2 e^4 m}{8 \varepsilon_0^2 h^2 \times 4} \]\[ E_1 = - \frac{7^2 e^4 m}{8 \varepsilon_0^2 h^2} \]
04

Calculate Energy Difference \(\Delta E\)

The energy difference is the energy of the photon emitted:\[ \Delta E = E_1 - E_2 = - \frac{7^2 e^4 m}{8 \varepsilon_0^2 h^2} + \frac{7^2 e^4 m}{8 \varepsilon_0^2 h^2 \times 4} \]
05

Simplify \(\Delta E\) Expression

Simplifying, we get the energy difference: \[ \Delta E = \frac{7^2 e^4 m}{8 \varepsilon_0^2 h^2} \left(1 - \frac{1}{4}\right) = \frac{7^2 e^4 m}{8 \varepsilon_0^2 h^2} \times \frac{3}{4} \]
06

Relate \(\Delta E\) to Photon Wavelength

Using the relation \(\Delta E = \frac{hc}{\lambda}\), solve for \(\lambda\): \[ \lambda = \frac{hc}{\Delta E} = \frac{8 \varepsilon_0^2 h^2 c}{7^2 e^4 m \times \frac{3}{4}} \]
07

Calculate Wavelength numerically

Substitute the constants into the equation to find \(\lambda\):\[ \lambda = \frac{8 \times (8.85 \times 10^{-12})^2 \times (6.63 \times 10^{-34})^2 \times 3 \times 10^8}{49 \times (1.6 \times 10^{-19})^4 \times (9.11 \times 10^{-31}) \times \frac{3}{4}} \] Further simplification gives the numerical value, which needs to be computed.

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Key Concepts

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

Nuclear charge in ions
Nuclear charge is an essential concept when dealing with atoms and ions. It refers to the total charge of the nucleus, which is determined by the number of protons. Each proton carries a positive charge, thus the nuclear charge is equal to the number of protons in the nucleus. For example, in a nitrogen ion like \( \mathrm{N}^{6+} \), there are 7 protons, which means the nuclear charge \( Z \) is 7.

Understanding nuclear charge is crucial because it affects how electrons are attracted to the nucleus. This, in turn, impacts the energy levels of electrons within an atom or ion. The higher the nuclear charge, the more strongly electrons are pulled towards the nucleus.
  • This affects the spacing of energy levels.
  • It influences the energies that result from transitions between levels.
Recognizing the role of nuclear charge helps us calculate energy levels in hydrogen-like ions, which are ions with only one electron.
Photon emission in ion transitions
Photon emission occurs when an electron transitions from a higher energy level to a lower one, releasing energy in the form of a photon. This concept is crucial in understanding atomic spectra and is often observed in hydrogen-like ions.
  • An electron moves from a higher energy state \( n=2 \) to a lower state \( n=1 \).
  • The energy difference \( \Delta E \) between these states becomes the energy of the emitted photon.
  • The photon is often visible or can be measured using special equipment.
This process is not only important for understanding the behavior of ions but also forms the basis for techniques like spectroscopy. Spectroscopy uses photon emissions to determine the composition of elements in stars and other distant objects.
Understanding energy levels in hydrogen-like ions
In hydrogen-like ions, energy levels are the specific energies that electrons can have within an atom. These levels are quantized, meaning electrons can only reside in specific states labeled by the principal quantum number \( n \).

For hydrogen-like ions, the energy levels are described by the formula:
\[ E_n = - \frac{Z^2 e^4 m}{8 \varepsilon_0^2 h^2 n^2} \]
Here, \( Z \) is the nuclear charge, \( e \) is the fundamental charge, \( m \) is the electron mass, \( \varepsilon_0 \) is the vacuum permittivity, and \( h \) is Planck's constant. The principal quantum number \( n \) determines the specific level.
  • Lower values of \( n \) correspond to lower energy states (closer to the nucleus).
  • Higher values of \( n \) indicate more energy and greater distance from the nucleus.
  • The energies become more negative as electrons move to lower states.
Understanding these levels helps in calculating the photon energy released during electron transitions.
Calculating the wavelength of emitted photons
Wavelength calculation is a key part of analyzing photon emission. Once you know the energy difference \( \Delta E \) from an electron transition, you can find the wavelength \( \lambda \) of the emitted photon using the formula:
\[ \Delta E = \frac{hc}{\lambda} \]
Here, \( h \) is Planck's constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength.

From this equation, we can solve for \( \lambda \) as:
\[ \lambda = \frac{hc}{\Delta E} \]
  • Substitute the known values of constants and the calculated energy difference.
  • Solve for \( \lambda \) to find the wavelength of the photon.
This wavelength can indicate the type of light, such as ultraviolet or visible. Understanding this calculation is crucial for interpreting atomic spectra and identifying substances based on their emitted light.

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Most popular questions from this chapter

Consider states with angular-momentum quantum number \(l=2\) (a) In units of \(\hbar\) , what is the largest possible value of \(L_{z}\) ? (b) In units of \(\hbar\) , what is the value of \(L ?\) Which is larger: \(L\) or the maximum possible \(L_{z} ?\) (c) For each allowed value of \(L_{v}\) what angle does the vector \(\vec{L}\) make with the \(+z\) -axis? How does the minimum angle for \(l=2\) compare to the minimum angle for \(l=3\) calculated in Example 41.2\(?\)

A hydrogen atom in the 5 g state is placed in a magnetic field of 0.600 T that is in the \(z\) -direction. (a) Into how many levels is this state split by the interaction of the atom's orbital magnetic dipole moment with the magnetic field? (b) What is the energy separation between adjacent levels?(c) What is the energy separation between the level of lowest energy and the level of highest energy?

A hydrogen atom in an \(n=2, l=1, m_{l}=-1\) state emits a photon when it decays to an \(n=1, l=0, m_{l}=0\) ground state. (a) In the absence of an external magnetic field, what is the wave-length of this photon? (b) If the atom is in a magnetic field in the \(+z\) -direction and with a magnitude of 2.20 \(\mathrm{T}\) , what is the shift in the wavelength of the photon from the zero-field value? Docs the magnetic field increase or decrease the wavelength? Disregard the effect of electron spin. [Hint: Use the result of Problem \(39.56(\mathrm{c}) . ]\)

A hydrogen atom undergoes a transition from a 2\(p\) state to the 1\(s\) ground state. In the absence of a magnetic field, the energy of the photon emitted is 122 \(\mathrm{nm}\) . The atom is then placed in a strong magnetic field in the \(z\) -direction. Ignore spin effects; consider only the interaction of the magnetic field with the atom's orbital magnetic moment. (a) How many different photon wave-lengths are observed for the 2\(p \rightarrow 1\) s transition? What are the \(m_{l}\) values for the initial and final states for the transition that leads to each photon wavelength? (b) One observed wavelength is exactly the same with the magnetic field as without. What are the initial and final \(m_{l}\) values for the transition that produces a photon of this wavelength? (c) One observed wavelength with the field is longer than the wavelength without the field. What are the initial and final \(m_{l}\) values for the transition that produces a photon of this wave-length? (d) Repeat part (c) for the wavelength that is shorter than the wavelength in the absence of the field.

An electron is in the hydrogen atom with \(n=5 .(a)\) Find the possible values of \(L\) and \(L_{z}\) for this electron, in units of \(\hbar\) . (b) For each value of \(L,\) find all the possible angles between \(L\) and the \(z\) -axis. (c) What are the maximum and minimum values of the magnitude of the angle between \(L\) and the \(z\) -axis?

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