Problem 46 - Chemistry The Central Science

Question

Place the following transitions of the hydrogen atom in order from shortest to longest wavelength of the photon emitted: \(n=5\) to \(n=3, n=4\) to \(n=2, n=7\) to \(n=4,\) and \(n=3\) to \(n=2\) .

Step-by-Step Solution

Answer

The order of the given transitions from shortest to longest wavelength of the photon emitted is: \(n=5\) to \(n=3\), \(n=7\) to \(n=4\), \(n=4\) to \(n=2\), and \(n=3\) to \(n=2\).

Step 1: Understanding the Rydberg formula for hydrogen atom

The Rydberg formula can be used to find the wavelength of the emitted photon during an electron transition in the hydrogen atom. The formula is given by: \[ \frac{1}{\lambda} = R_{H} \left( \frac{1}{n_{1}^2} - \frac{1}{n_{2}^2} \right) \] where \(\lambda\) is the wavelength, \(R_{H}\) is the Rydberg constant for hydrogen atom (\(1.097 \times 10^7 m^{-1}\)), \(n_{1}\) is the lower energy level, and \(n_{2}\) is the higher energy level.

Step 2: Calculate the wavelength for each transition

We'll plug in the given values for \(n_1\) and \(n_2\) in the Rydberg formula to find the wavelength for each transition: 1) \(n=5 \rightarrow n=3\): \[ \frac{1}{\lambda_{1}} = R_H\left(\frac{1}{3^2} - \frac{1}{5^2}\right) \] 2) \(n=4 \rightarrow n=2\): \[ \frac{1}{\lambda_{2}} = R_H\left(\frac{1}{2^2} - \frac{1}{4^2}\right) \] 3) \(n=7 \rightarrow n=4\): \[ \frac{1}{\lambda_{3}} = R_H\left(\frac{1}{4^2} - \frac{1}{7^2}\right) \] 4) \(n=3 \rightarrow n=2\): \[ \frac{1}{\lambda_{4}} = R_H\left(\frac{1}{2^2} - \frac{1}{3^2}\right) \]

Step 3: Solving for the wavelengths

Calculate the value of \(\lambda_i\) for each transition: 1) \(\lambda_{1} = 1.6 \times 10^{-7}\,m\) 2) \(\lambda_{2} = 4.86 \times 10^{-7}\,m\) 3) \(\lambda_{3} = 2.97 \times 10^{-7}\,m\) 4) \(\lambda_{4} = 6.56 \times 10^{-7}\,m\)

Step 4: Arrange the wavelengths from shortest to longest

With the calculated values for each wavelength, we can now arrange them in the required order: 1) \(1.6 \times 10^{-7}\,m\) (\(n=5 \rightarrow n=3\)) 2) \(2.97\times 10^{-7}\,m\) (\(n=7 \rightarrow n=4\)) 3) \(4.86 \times 10^{-7}\,m\) (\(n=4 \rightarrow n=2\)) 4) \(6.56 \times 10^{-7}\,m\) (\(n=3 \rightarrow n=2\)) So, the order of the given transitions from shortest to longest wavelength of the photon emitted is: \(n=5\) to \(n=3\), \(n=7\) to \(n=4\), \(n=4\) to \(n=2\), and \(n=3\) to \(n=2\).

Key Concepts

Hydrogen Atom Transitions

In hydrogen atoms, electrons transition between different energy levels. These energy levels are denoted by the principal quantum number, \(n\). Each transition corresponds to an electron moving from a higher energy state \(n_2\) to a lower one \(n_1\). The energy difference between these levels releases or absorbs energy as a photon. The Rydberg formula is commonly used to calculate the wavelengths of emitted or absorbed photons during these transitions. Using this formula, scientists can predict the wavelengths of light emitted when electrons jump between energy levels. This is crucial for understanding the spectral lines of hydrogen, which serve as a great tool for investigations in astrophysics and quantum mechanics.

• The larger the gap between energy levels, the more energy is released and the shorter the emitted photon's wavelength.
• When the electron transitions downward, it emits energy and the wavelength of the photon emitted can be calculated using the Rydberg formula.

Photon Emission

When electrons transition to lower energy levels, they emit energy in the form of photons. This process is known as photon emission. The photon's wavelength is inversely related to the energy difference between the two levels. That means a larger energy difference corresponds to a shorter wavelength. The emitted photon is a result of the electron's loss of potential energy. In the case of hydrogen, the color of the emitted light can range from ultraviolet to visible light, depending on the electron drop's energy difference.

• Shorter wavelength photons possess higher energy and are often visible as blue or violet light.
• Longer wavelength photons have lower energy and may appear red or infrared.

Frequency and energy of the emitted photon can also be calculated with the help of the formula \(E=hf\), where \(E\) is energy, \(h\) is Planck's constant, and \(f\) is frequency.

Electron Transition

Electron transitions within an atom involve the movement of electrons from one quantum state to another. These transitions can happen either by absorbing energy to move to a higher level or releasing energy to move to a lower one. In hydrogen, each transition leads to the emission or absorption of photons with specific energies and wavelengths. The Rydberg formula is essential for calculating these energies based on the transition levels

• The transitions that involve larger differences between energy levels result in the emission of higher energy photons.
• These transitions allow scientists to study atomic structure and the behavior of electrons.

Understanding electron transitions helps in analyzing atomic spectra, which are characteristic of each element, aiding in chemical analysis and identification.