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Chapter 2: Problem 42

According to Bohr's theory, the electronic energy of hydrogen atom in the \(n^{\text {ti }}\) Bohr's orbit is given \(\operatorname{by} E_{n}=\frac{-21.76 \times 10^{-19}}{n^{2}} \mathrm{~J}\). Calculate the longest wavelength of light that will be needed to remove an electron from the third Bohr orbit of the \(\mathrm{He}^{+}\)ion.

Short Answer

Expert verified
The longest wavelength needed is approximately 206 nm.

Step by step solution

01

Understand the Energy Formula

The electronic energy of a hydrogen-like atom in the nth orbit is given by the formula \( E_n = \frac{-21.76 \times 10^{-19}}{n^2} \) J. Since we are dealing with an \( \text{He}^+ \) ion, we need to modify this formula to account for the nuclear charge \( Z = 2 \). The energy of the \( n^{th} \) orbit for a hydrogen-like ion is \( E_n = \frac{-21.76 \times 10^{-19} \cdot Z^2}{n^2} = \frac{-21.76 \times 10^{-19} \cdot 4}{n^2} \) J.
02

Calculate the Energy of the 3rd Orbital

Substitute \( n = 3 \) and \( Z = 2 \) into the modified energy formula: \[ E_3 = \frac{-21.76 \times 10^{-19} \cdot 4}{3^2} \]. Calculate the energy to find \( E_3 = \frac{-87.04 \times 10^{-19}}{9} \approx -9.67 \times 10^{-19} \) J.
03

Determine the Ionization Energy

The ionization energy is the energy needed to remove the electron from the atom, which corresponds to moving the electron from \( n = 3 \) to \( n = \infty \), or essentially zero energy. Thus, \( E_{ionization} = 0 - E_3 = 9.67 \times 10^{-19} \) J.
04

Convert Energy to Wavelength

Use the relation between energy and wavelength: \( E = \frac{hc}{\lambda} \), where \( h = 6.626 \times 10^{-34} \) Js is Planck's constant and \( c = 3 \times 10^8 \) m/s is the speed of light. Rearrange to solve for wavelength: \( \lambda = \frac{hc}{E} \).
05

Calculate Longest Wavelength

Substitute \( h = 6.626 \times 10^{-34} \) Js, \( c = 3 \times 10^8 \) m/s, and \( E = 9.67 \times 10^{-19} \) J into the wavelength formula: \[ \lambda = \frac{6.626 \times 10^{-34} \cdot 3 \times 10^8}{9.67 \times 10^{-19}} \]. Calculate \( \lambda \approx 2.06 \times 10^{-7} \) m or 206 nm.

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Key Concepts

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

Hydrogen-like Ions
Hydrogen-like ions are ions that have only one electron, similar to the hydrogen atom. These include singly-charged ions like He+, Li2+, and so on.
These ions are characterized by having a nuclear charge that can be different from hydrogen's simple single proton, which means they have a different atomic number, Z.
This leads to different energy levels compared to a typical hydrogen atom, increasing complexity for calculations based on Bohr's theory.
  • One-electron system: Hydrogen-like ions have a single electron that orbits around a nucleus with more than one proton.
  • Modifications in Bohr's Formula: The energy formula for Bohr's orbit must be adjusted by including the nuclear charge Z. For hydrogen-like ions, the formula is \( E_n = \frac{-21.76 \times 10^{-19} Z^2}{n^2} \) J, where Z is the atomic number.
  • Importance in calculations: Understanding hydrogen-like ions aids in calculating energy levels and ionization energies for systems beyond simple hydrogen atoms.
Recognizing these ions is crucial when dealing with atomic transition energies, as seen in many branches of physics and chemistry.
Ionization Energy
Ionization energy is the energy required to remove an electron from an atom or ion in its gaseous state. For hydrogen-like ions, this concept expands since additional energy is required due to the increased nuclear charge. This process involves an electron transitioning from a bounded state (such as energy level n=3) to a free state (n=鈭�).
  • Bohr's Concept: According to Bohr's theory, calculating the ionization energy involves evaluating the energy needed to move an electron from a specific orbit to the point where it no longer feels the attractive force of the nucleus.
  • Energy Transition: For the He+ ion with modified energy levels, the ionization energy from the third orbit is calculated by determining the energy difference from \( n = 3 \) to \( n = \infty \).\( E_{ionization} = 0 - E_3 = 9.67 \times 10^{-19} \) J.
  • Significance: Understanding ionization energy is fundamental in fields like quantum mechanics, spectroscopy, and chemistry, where precise electronic transitions are pivotal.
Acknowledging ionization energy helps us understand how electrons behave under the influence of varying nuclear charges, which is fundamental to atomic theory.
Wavelength Calculation
Calculating the wavelength of energy transitions is essential to understanding the electromagnetic spectrum's details. According to the energy-wavelength relationship, the energy of a photon can be linked to its wavelength through the formula \( E = \frac{hc}{\lambda} \), where h is Planck's constant and c is the speed of light.
  • Application of the formula: For transitions involving ionization energy, this relationship helps determine the wavelength of light associated with the energy required to ionize an electron.
  • Conducting the calculation: Given an energy \( E = 9.67 \times 10^{-19} \) J for ionization, the wavelength \( \lambda \) is calculated using the formula \( \lambda = \frac{hc}{E} \).Substituting the known constants and calculated energy yields \( \lambda \approx 2.06 \times 10^{-7} \) m or 206 nm.
  • Interpreting Results: The wavelength corresponds to the longest wavelength of light necessary for ionizing the electron from the given energy level, particularly crucial in spectrometric applications.
Wavelength calculations not only connect quantum mechanics with electromagnetic theory but also enable detailed analysis of light-matter interactions.

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Most popular questions from this chapter

An isotone of \({ }_{32}^{76} \mathrm{Ge}\) is : (a) \({ }_{32}^{77} \mathrm{Ge}\) (b) \({ }_{33}^{77} \mathrm{As}\) (c) \({ }_{34}^{77} \mathrm{Se}\) (d) \({ }_{34}^{78} \mathrm{Se}\)

Find the velocity \(\left(\mathrm{ms}^{-1}\right)\) of electron in first Bohr's orbit of radius \(a_{0}\). Also find the de Broglie's wavelength (in \(\mathrm{m}\) ). Find the orbital angular momentum of \(2 p\) orbital of hydrogen atom in units of \(h / 2 \pi\).

Given that the abundances of isotopes \({ }^{54} \mathrm{Fe},{ }^{56} \mathrm{Fe}\) and \({ }^{\mathrm{s} 7} \mathrm{Fe}\) are \(5 \%, 90 \%\) and \(5 \%\), respectively, the atomic mass of \(\mathrm{Fe}\) is (a) \(55.85\) (b) \(55.95\) (c) \(55.75\) (d) \(56.05\)

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The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is \(\left[a_{0}\right.\) is Bohr radius] : (a) \(\frac{h^{2}}{4 \pi^{2} m a_{0}^{2}}\) (b) \(\frac{h^{2}}{16 \pi^{2} m a_{0}^{2}}\) (c) \(\frac{h^{2}}{32 \pi^{2} m a_{0}^{2}}\) (d) \(\frac{h^{2}}{64 \pi^{2} m a_{0}^{2}}\)
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