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Chapter 6: Problem 39

(a) Using Equation \(6.5,\) calculate the energy of an electron in the hydrogen atom when \(n=2\) and when \(n=6 .\) Calculate the wavelength of the radiation released when an electron moves from \(n=6\) to \(n=2 .\) (b) Is this line in the visible region of the electromagnetic spectrum? If so, what color is it?

Short Answer

Expert verified
The energy of the electron at energy levels n=2 is -3.4 eV and at n=6 is -0.378 eV. The energy difference between the two levels is -3.022 eV, which corresponds to a wavelength of 407 nm when the electron moves from n=6 to n=2. This wavelength is in the visible region of the electromagnetic spectrum and corresponds to the color violet.

Step by step solution

01

Calculate the energy levels of the electron at n=2 and n=6

We are given that Equation 6.5 is used to calculate the energy levels of an electron in a hydrogen atom. The most commonly used formula is: \( E_n = -\dfrac{13.6 \text{ eV}}{n^2} \) Where, \(E_n\) is the energy of the electron at the energy level n. First, calculate the energy level for n=2: \( E_2 = -\dfrac{13.6 \text{ eV}}{2^2} = -\dfrac{13.6 \text{ eV}}{4} = -3.4 \text{ eV}\) Now, calculate the energy level for n=6: \( E_6 = -\dfrac{13.6 \text{ eV}}{6^2} = -\dfrac{13.6 \text{ eV}}{36} = -0.378 \text{ eV}\)
02

Determine the energy difference between the two energy levels

Now that we have the energy levels for n=2 and n=6, we can calculate the energy difference as the electron moves from n=6 to n=2: \( \Delta E = E_2 - E_6 = -3.4 \text{ eV} - (-0.378 \text{ eV}) = -3.022 \text{ eV}\)
03

Calculate the wavelength of the radiation released

To calculate the wavelength of the radiation released, we can use the energy-wavelength relationship: \( \lambda = \dfrac{hc}{E}\) Where, h is Planck's constant (4.1357 x 10^-15 eV s), c is the speed of light (2.99792 x 10^8 m/s), and E is the energy difference. Convert the energy difference to Joules: \( \Delta E = -3.022 \text{ eV} \cdot \dfrac{1.60218 \times 10^{-19} \text{ J}}{1 \text{ eV}} = -4.839 \times 10^{-19} \text{ J}\) Now, calculate the wavelength: \( \lambda = \dfrac{4.1357 \times 10^{-15} \text{ eV s} \cdot 2.99792 \times 10^8 \text{ m/s}}{-4.839 \times 10^{-19} \text{ J}} = 4.07 \times 10^{-7} \text{ m} \)
04

Check if the wavelength falls within the visible spectrum and determine its color if applicable

The visible spectrum ranges from about 400 nm to 700 nm. To check if the calculated wavelength falls within this range, convert the wavelength to nanometers: \( \lambda = 4.07 \times 10^{-7} \text{ m} \cdot \dfrac{10^9 \text{ nm}}{1 \text{ m}} = 407 \text{ nm}\) The calculated wavelength (407 nm) is within the visible spectrum. Based on the visible color wavelengths, this wavelength corresponds to the color violet. In conclusion, the radiation released as the electron moves from the energy level n=6 to n=2 has a wavelength of 407 nm, which corresponds to violet in the visible spectrum.

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Key Concepts

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

Energy Levels in Atoms
At the heart of quantum mechanics is the concept of discrete energy levels within atoms. These levels represent the quantized energies that an electron in an atom can have. In the hydrogen atom, for example, these levels are defined by the principal quantum number, denoted as 'n'. Each level is inversely proportional to the square of 'n', which means the higher the 'n', the lower the absolute value of energy (closer to zero), as seen in the provided exercise.

Imagine the energy levels as rungs on a ladder that an electron can 'climb'. The electron can only reside on these rungs and nowhere in between. This quantization is elegantly described by the formula:
\( E_n = -\dfrac{13.6 \text{ eV}}{n^2} \)

If we visualize the atom in a simplified way, the electron orbits closer to the nucleus at lower energy levels (small 'n'), and further away at higher levels (large 'n'). When examining an electron making a transition from a higher level to a lower level, it emits energy in the form of electromagnetic radiation. This energy difference accounts for the specific properties of the emitted photon, such as its wavelength, which we can calculate.
Wavelength Calculation
Deciphering the wavelengths of photons released during electron transitions is like tuning into different radio frequencies. The wavelength calculation bridges the gap between energy changes within the atom and the type of light we can observe. The energy-wavelength relationship, explained in the solution, reveals how the energy of a photon determines its wavelength.

The formula for wavelength calculation takes into account the energy released during the transition (\( \Delta E \)), Planck's constant (\( h \)), and the speed of light (\( c \)):
\( \lambda = \dfrac{hc}{E} \)

For our hydrogen atom scenario, where the electron transitions from \( n=6 \) to \( n=2 \), by working through the math with the given constants, we pinpoint the exact wavelength of the emitted light. This precise calculation can lead us to identify whether the light emitted is visible to the naked eye and, if so, what color it corresponds to within the visible spectrum.
Visible Spectrum Colors
The visible spectrum is a tiny but vibrant fraction of the electromagnetic spectrum, sandwiched between ultraviolet and infrared light, which our eyes can detect. With wavelengths ranging approximately from 400 nm (violet) to 700 nm (red), each color we see corresponds to a specific wavelength within this range.

As calculated in the exercise, a photon with a wavelength of 407 nm falls within the violet end of this spectrum, gracing us with the vision of a deep purple glow. The shorter the wavelength, the closer the color is to violet, and as wavelengths increase, the colors shift through the spectrum towards red. This continuous 'rainbow' enables us to enjoy the full palette of colors present in nature, including the specific color of light emitted when an electron transitions between different energy levels in a hydrogen atom.

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Most popular questions from this chapter

Molybdenum metal must absorb radiation with a minimum frequency of \(1.09 \times 10^{15} \mathrm{s}^{-1}\) before it can eject an electron from its surface via the photoelectric effect. (a) What is the minimum energy needed to eject an electron? (b) What wavelength of radiation will provide a photon of this energy? (c) If molybdenum is irradiated with light of wavelength of \(120 \mathrm{nm},\) what is the maximum possible kinetic energy of the emitted electrons?

For orbitals that are symmetric but not spherical, the contour representations (as in Figures 6.23 and 6.24 ) suggest where nodal planes exist (that is, where the electron density is zero). For example, the \(p_{x}\) orbital has a node wherever \(x=0\) . This equation is satisfied by all points on the \(y z\) plane, so this plane is called a nodal plane of the \(p_{x}\) orbital. (a) Determine the nodal plane of the \(p_{z}\) orbital. (b) What are the two nodal planes of the \(d_{x y}\) orbital? (c) What are the two nodal planes of the \(d_{x^{2}-y^{2}}\) orbital?

The energy from radiation can be used to cause the rupture of chemical bonds. A minimum energy of 242 \(\mathrm{kJ} / \mathrm{mol}\) is required to break the chlorine-chlorine bond in \(\mathrm{Cl}_{2} .\) What is the longest wavelength of radiation that possesses the necessary energy to break the bond? What type of electromagnetic radiation is this?

Bohr's model can be used for hydrogen-like ions-ions that have only one electron, such as \(\mathrm{He}^{+}\) and \(\mathrm{Li}^{2+} .\) (a) Why is the Bohr model applicable to He \(^{+}\) ions but not to neutral He atoms? (b) The ground-state energies of \(\mathrm{H}, \mathrm{He}^{+},\) and \(\mathrm{Li}^{2+}\) are tabulated as follows: $$ \begin{array}{l}{\text { Atom or ion } \quad \quad\quad\quad\quad\quad \mathrm{H} \quad\quad\quad\quad\quad\quad \text { He }^{+} \quad\quad\quad\quad\quad\quad\quad \mathrm{Li}^{2+}} \\ {\text { Ground- state }\quad-2.18 \times 10^{-18} \mathrm{J}\quad-8.72 \times 10^{-18} \mathrm{J}\quad-1.96 \times 10^{-17} \mathrm{J}} \\ {\text { energy }}\end{array} $$ By examining these numbers, propose a relationship between the ground-state energy of hydrogen-like systems and the nuclear charge, \(Z .(\mathbf{c})\) Use the relationship you derive in part (b) to predict the ground-state energy of the \(\mathrm{C}^{5+}\) ion.

An experiment called the Stern-Gerlach experiment helped establish the existence of electron spin. In this experiment, a beam of silver atoms is passed through a magnetic field, which deflects half of the silver atoms in one direction and half in the opposite direction. The separation between the two beams increases as the strength of the magnetic field increases. (a) What is the electron configuration for a silver atom? (b) Would this experiment work for a beam of cadmium (Cd) atoms? (c) Would this experiment work for a beam of fluorine (F) atoms?
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