Question

When the excited electron in a hydrogen atom falls from \(n=5\) to \(n=2,\) a photon of blue light is emitted. If an excited electron in He \(^{+}\) falls from \(n=4,\) to which energy level must it fall so that a similar blue light (as with the hydrogen) is emitted? Prove it. (See Exercise \(174 . )\)

Short answer

The excited electron in the helium ion must fall to the energy level approximately equal to \(+\sqrt{45}\) (≈ 6.71) to emit a similar blue light as when an electron falls from n=5 to n=2 in the hydrogen atom. This is derived using the Rydberg formula for the energy difference, setting up the equation for the ratio of energy differences, and solving for n.

Step-by-step solution

Calculate the energy difference for hydrogen emission

First, we will find the energy difference when an electron falls from n=5 to n=2 in a hydrogen atom. To do this, we'll use the Rydberg formula for the energy difference: \(ΔE_n = -13.6 eV * \frac{Z^2}{n^2}\) Where: - \(ΔE_n\) is the energy difference - \(13.6 eV\) is the Rydberg constant for hydrogen (approximated value) - Z is the atomic number, which is 1 for hydrogen - n is the energy level to which the electron falls. The energy difference for hydrogen atom is: \(ΔE_{H} = -13.6 eV * \frac{1^2}{5^2} + 13.6 eV * \frac{1^2}{2^2}\)

Calculate the energy difference for the helium ion emission

Now, we will calculate the energy difference for helium ion emission using the Rydberg formula, keeping in mind that Z=2 for Helium: \(ΔE_{He} = -13.6 eV * \frac{2^2}{4^2} + 13.6 eV * \frac{2^2}{n^2}\) Where n is the energy level to which the electron falls in the helium ion.

Set up the ratio of the energy differences and solve for n

We know that the ratio of the energy differences for hydrogen and helium ion should be equal. So, we set up the equation and solve for n: \(\frac{ΔE_H}{ΔE_{He}} = 1\) Substitute the energy difference expressions from steps 1 and 2: \(\frac{-13.6 * \frac{1^2}{5^2} + 13.6 * \frac{1^2}{2^2}}{-13.6 * \frac{2^2}{4^2} + 13.6 * \frac{2^2}{n^2}} = 1\) Now, we can cancel out \(13.6 eV\) on both sides and solve for n: \(\frac{\frac{1^2}{5^2} - \frac{1^2}{2^2}}{\frac{2^2}{4^2} - \frac{2^2}{n^2}} = 1\)

Solve the equation for n

Next, we need to solve the equation obtained in step 3 for n: \(n^2 = 45\) Square root of both sides: \(n = \pm\sqrt{45}\) Since only the positive value of n is meaningful in this context, we consider n = \(+\sqrt{45}\). So the excited electron in the helium ion must fall to the energy level approximately equal to \(+\sqrt{45}\) (≈ 6.71) to emit a similar blue light as when an electron falls from n=5 to n=2 in the hydrogen atom.

Key concepts

Energy Levels

In an atom, electrons orbit the nucleus in distinct energy levels or shells. These levels, commonly referred to by their principal quantum number, n, represent the allowed energies for electrons within the atom. For instance, in the hydrogen atom, these levels are denoted by n=1,2,3, etc. The lower the value of n, the closer the electron is to the nucleus, and the less energy it has. Conversely, as n increases, electrons possess higher energy and are situated further from the nucleus.

Transitions between these levels, such as when an electron falls from a higher to a lower level, involve the absorption or emission of energy in the form of a photon. Understanding these energy levels is crucial for solving problems involving electron transitions such as photon emission.

Photon Emission

Photon emission occurs when an electron transitions between two different energy levels in an atom. When an electron falls from a higher energy level to a lower one, the energy difference between these levels is released as a photon. The wavelength and color of the emitted photon depend on the specific energy change involved.

Using the Rydberg formula, one can calculate this energy difference and hence determine the emitted photon's wavelength. The formula is often expressed as: \[ ΔE = -13.6 \, eV \left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right)\]where \(n_f\) and \(n_i\) are the final and initial energy levels, respectively. This theoretical framework helps in predicting the characteristics of light emitted by various atomic transitions.

Hydrogen Atom

The hydrogen atom, consisting of one electron orbiting a single proton, serves as a fundamental model for studying atomic behavior. It was crucial in developing quantum mechanics and understanding atomic structure. The hydrogen atom's simplicity allows detailed calculations of its energy levels using the Rydberg formula, which captures the behavior of the sole electron around the nucleus.

The Rydberg constant (13.6 eV for hydrogen) is derived from hydrogen's properties. Employing this constant, the energy associated with electron transitions between different levels in hydrogen can be calculated, revealing its emission and absorption spectra. Such deep insights into the hydrogen atom have paved the way for understanding more complex elements and ions.

Helium Ion

The helium ion (He\(^+\)), formed when a helium atom loses one of its two electrons, is an essential example of a hydrogen-like ion but with a significant difference. Unlike hydrogen, helium ion has two protons in its nucleus, thereby doubling the nuclear charge (Z=2). This increased nuclear charge makes the helium ion's energy levels more tightly bound compared to hydrogen.

Adjusting the Rydberg formula for the helium ion involves replacing the atomic number Z in the formula. \[ ΔE = -13.6 \, eV \times Z^{2} \left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right)\]For helium ion, with Z=2, this adjustment captures how the energy differences scale due to the higher proton count. Thus, while the helium ion and hydrogen share some similarities, the enhanced nuclear charge profoundly affects energy levels and emitted photon characteristics.