Problem 16 - College Physics

Question

\(\bullet\) (a) How much energy is needed to ionize a hydrogen atom that is in the \(n=4\) state? (b) What would be the wavelength of a photon emitted by a hydrogen atom in a transition from the \(n=4\) state to the \(n=2\) state?

Step-by-Step Solution

Answer

(a) 0.85 eV is needed for ionization from \(n=4\). (b) The emitted photon's wavelength is 486 nm.

Step 1: Understand the Ionization Energy Formula

The ionization energy of an atom in a given state is the energy required to completely remove an electron from that state. For hydrogen, this energy can be calculated using the formula: \[ E_n = - \frac{13.6\text{ eV}}{n^2} \]where \(n\) is the principal quantum number of the initial state. The total energy needed to ionize the atom from state \(n\) is the absolute value of \(E_n\).

Step 2: Calculate Energy for Ionization from n=4

Substitute \(n=4\) into the ionization energy formula:\[ E_4 = - \frac{13.6\text{ eV}}{4^2} = - \frac{13.6}{16}\text{ eV} = -0.85\text{ eV} \]Thus, the energy needed is the absolute value: \[ |E_4| = 0.85\text{ eV} \]

Step 3: Photon Emission Energy Calculation

For part (b), we need to calculate the energy difference between the \(n=4\) and \(n=2\) states, which is the energy of the emitted photon. Use:\[ \Delta E = E_2 - E_4 \]First, calculate \(E_2\):\[ E_2 = - \frac{13.6\text{ eV}}{2^2} = -3.4\text{ eV} \]Then, calculate \(\Delta E\):\[ \Delta E = -3.4\text{ eV} - (-0.85\text{ eV}) = -3.4 + 0.85 = -2.55\text{ eV} \]

Step 4: Convert Energy to Wavelength

Convert the energy of the emitted photon to wavelength using the relation:\[ \lambda = \frac{hc}{|\Delta E|} \]where \(h = 6.626 \times 10^{-34}\text{ Js}\) is Planck's constant, \(c = 3 \times 10^8\text{ m/s}\) is the speed of light, and \(|\Delta E| = 2.55\text{ eV}\). Convert electronvolts to joules (1 eV = \(1.602 \times 10^{-19}\text{ J}\)):\[ |\Delta E| = 2.55 \times 1.602 \times 10^{-19}\text{ J} = 4.0851 \times 10^{-19}\text{ J} \]Now calculate \(\lambda\):\[ \lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4.0851 \times 10^{-19}} = 486 \text{ nm} \]

Step 5: Summary of Calculations

(a) The energy needed to ionize a hydrogen atom from the \(n=4\) state is 0.85 eV. (b) The wavelength of a photon emitted when a hydrogen atom transitions from the \(n=4\) to the \(n=2\) state is 486 nm.

Key Concepts

Principal Quantum Number

The principal quantum number, often symbolized by the letter \( n \), plays a crucial role in the quantum mechanical model of an atom. It reveals the energy level of an electron within an atom. Imagine it as a kind of address for an electron's location with respect to its distance from the nucleus. The higher the principal quantum number, the further the electron likely is from the nucleus.

• The principal quantum number is an integer starting from 1 (\( n = 1, 2, 3, \ldots \)).
• Each value of \( n \) represents a "shell" where electrons are found.
• As \( n \) increases, so does the electron's energy and the space it occupies.

For a hydrogen atom, which has only one electron, the principal quantum number directly defines the energy state of this electron. When \( n = 4 \), the electron resides in a higher energy level, making it less tightly bound to the nucleus compared to lower \( n \) levels.

Energy Level Transitions

Energy level transitions occur when an electron moves from one energy level to another within an atom. This movement is usually accompanied by the absorption or emission of energy. In hydrogen, a simple change in the principal quantum number, such as from \( n = 4 \) to \( n = 2 \), is an example of a transition.

• When an electron falls to a lower energy level, such as from \( n = 4 \) to \( n = 2 \), it releases energy.
• The energy released in such a transition is equal to the difference in energy between the two levels.
• Conversely, energy must be absorbed for an electron to move to a higher energy level.

Calculating these energy differences requires understanding the specific energies of the \( n \)-levels according to the formula \( E_n = - \frac{13.6 \text{ eV}}{n^2} \) for hydrogen. Using this formula, we can find out how much energy the electron emits as it transitions between two states, as highlighted in the problem.

Photon Emission

Photon emission is an exciting process where an atom, like hydrogen, loses energy due to an electron transitioning to a lower energy level. This energy loss manifests as a photon, a type of light particle.

• The emitted photon carries energy equivalent to the energy difference between the two levels involved in the transition.
• This energy is directly linked to the frequency (\( u \)) and wavelength (\( \lambda \)) of the emitted light: \( E = hu = \frac{hc}{\lambda} \).
• The formula \( \lambda = \frac{hc}{|\Delta E|} \) converts the energy of the emitted photon into its wavelength.

Through this process, different transitions in hydrogen result in various wavelengths appearing, forming a spectrum that identifies particular energy changes. For example, in this particular exercise, the transition from \( n=4 \) to \( n=2 \) results in a photon with a wavelength of 486 nm, a visible light in the Balmer series of hydrogen spectra.