Problem 39 - Chemistry The Central Science

Question

(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?

Step-by-Step Solution

Answer

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 1: 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: \( E_6 = -\dfrac{13.6 \text{ eV}}{6^2} = -\dfrac{13.6 \text{ eV}}{36} = -0.378 \text{ eV}\)

Step 2: 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: \( \Delta E = E_2 - E_6 = -3.4 \text{ eV} - (-0.378 \text{ eV}) = -3.022 \text{ eV}\)

Step 3: 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: \( \lambda = \dfrac{4.1357 \times 10^{-15} \text{ eV s} \cdot 2.99792 \times 10^8 \text{ m/s}}{-4.839 \times 10^{-19} \text{ J}} = 4.07 \times 10^{-7} \text{ m} \)

Step 4: 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.

Key Concepts

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.