Question

(a) Using Equation 6.5 , calculate the energy of an electron in the hydrogen atom when \(n=3\) and when \(n=6\). Calculate the wavelength of the radiation released when an electron moves from \(n=6\) to \(n=3 .(\mathbf{b})\) Is this line in the visible region of the electromagnetic spectrum?

Short answer

The electron energy is \(-1.51\) eV for \(n=3\) and \(-0.378\) eV for \(n=6\). The energy difference between these two energy levels is \(1.132\) eV. The wavelength of the radiation released when the electron moves from \(n=6\) to \(n=3\) is approximately \(1.095 \times 10^{-6} \text{ m}\) or \(1095 \text{ nm}\), which is not within the visible region of the electromagnetic spectrum (380 nm to 750 nm).

Step-by-step solution

Calculate the electron energy for \(n=3\) and \(n=6\) using Equation 6.5

To find the energy of the electron for \(n=3\) and \(n=6\), we will use Equation 6.5: \[E_n = -\frac{13.6 \text{ eV}}{n^2}\] For \(n=3\): \[E_3 = -\frac{13.6 \text{ eV}}{(3)^2} = -\frac{13.6 \text{ eV}}{9} = -1.51 \text{ eV}\] For \(n=6\): \[E_6 = -\frac{13.6 \text{ eV}}{(6)^2} = -\frac{13.6 \text{ eV}}{36} = -0.378 \text{ eV}\] The electron energy is \(-1.51\) eV for \(n=3\) and \(-0.378\) eV for \(n=6\).

Calculate the energy difference between the two energy levels

Now, we need to find the energy difference between \(n=6\) and \(n=3\). For this, simply subtract the energy at \(n=3\) from the energy at \(n=6\): \[\Delta E = E_6 - E_3 = (-0.378 - (-1.51)) \text{ eV} = 1.132 \text{ eV}\] The energy difference between these two energy levels is \(1.132\) eV.

Calculate the wavelength of the radiation released

To calculate the wavelength of the radiation released, we will use the energy difference and the Planck's constant (\(h = 4.14 \times 10^{-15} \text{ eV s}\)): \[\lambda = \frac{hc}{\Delta E}\] First, convert the energy difference to Joules: \[\Delta E = 1.132 \text{ eV} \times \frac{1.6 \times 10^{-19} \text{ J}}{1 \text{ eV}} = 1.8112 \times 10^{-19} \text{ J}\] Now, calculate the wavelength: \[\lambda = \frac{(6.63 \times 10^{-34} \text{ J s})(3 \times 10^8 \text{ m/s})}{1.8112 \times 10^{-19} \text{ J}} = 1.095 \times 10^{-6} \text{ m}\] The wavelength of the radiation released is approximately \(1.095 \times 10^{-6} \text{ m}\) or \(1095 \text{ nm}\).

Determine if the wavelength is within the visible spectrum

The visible spectrum ranges from approximately 380 nm to 750 nm. Since the calculated wavelength is approximately \(1095 \text{ nm}\), it is not within the visible range. Therefore, the radiation released when the electron moves from \(n=6\) to \(n=3\) is not within the visible region of the electromagnetic spectrum.

Key concepts

Energy Levels in Hydrogen Atom

The hydrogen atom is a simple system that provides deep insights into how electrons occupy different energy levels. These levels are associated with the principal quantum number, denoted as \( n \). Each value of \( n \) represents a specific energy level, also known as an electron orbit. The energy levels are not continuous but quantized, meaning electrons can only exist in certain fixed levels. This quantization leads to distinct energy values for each possible orbit an electron might take.

For the hydrogen atom, the energy of an electron is given by the equation: \[ E_n = -\frac{13.6 \text{ eV}}{n^2} \] where \( 13.6 \text{ eV} \) is the ionization energy of hydrogen, and \( n \) is the principal quantum number.

As \( n \) increases, the absolute value of energy \(|E_n|\) decreases, meaning the electron is less tightly bound to the nucleus. This makes sense considering that higher energy levels are further away from the nucleus. In our exercise, when \( n = 3 \) and \( n = 6 \), the energy values calculated were \(-1.51 \text{ eV}\) and \(-0.378 \text{ eV}\) respectively. This shows the electron is more stable at \( n = 3 \) because it is closer to the nucleus, compared to at \( n = 6 \).

Wavelength of Electromagnetic Radiation

When an electron transitions between energy levels, it either absorbs or emits energy. This energy change manifests as electromagnetic radiation. The wavelength of this radiation depends on the energy difference between the initial and final levels. To calculate the wavelength of the emitted or absorbed radiation, use the equation: \[ \lambda = \frac{hc}{\Delta E} \] where \( h \) is Planck's constant \( (6.63 \times 10^{-34} \text{ J s}) \), \( c \) is the speed of light \( (3 \times 10^8 \text{ m/s}) \), and \( \Delta E \) is the energy difference between the two levels.

In our problem, the energy difference \( \Delta E \) between \( n = 6 \) and \( n = 3 \) was calculated as \( 1.132 \text{ eV} \), which is equivalent to \( 1.8112 \times 10^{-19} \text{ J} \). Inserting these values into the wavelength formula gives us a wavelength of approximately \( 1095 \text{ nm} \). This wavelength corresponds to the infrared region of the electromagnetic spectrum, as it is much longer than wavelengths in the visible region.

Visible Region of Electromagnetic Spectrum

The electromagnetic spectrum encompasses all types of electromagnetic radiation, ranging from gamma rays to radio waves, each distinguished by a particular wavelength. The visible region is a small portion of this spectrum, typically ranging from approximately 380 nm to 750 nm. This range is what the average human eye can perceive as visible light.

Different colors in visible light are associated with different wavelengths:

• Violet light is near 380 nm.
• Green light is around 550 nm.
• Red light extends to about 750 nm.

When an electron transitions and emits radiation, the wavelength determines whether the light is visible or falls into infrared, ultraviolet, or other sections of the spectrum.

For our example, the transition from \( n = 6 \) to \( n = 3 \) resulted in a wavelength of approximately \( 1095 \text{ nm} \), which lies in the infrared region, not visible to the human eye. Such transitions and emissions are crucial for understanding phenomena and applications like spectroscopy, where invisible light is used to gain information about material structures.