Problem 78 - Fundamentals of Physics

Question

Consider an atom with two closely spaced excited states \(A\) and \(B\). If the atom jumps to ground state from \(A\) or from \(B\), it emits a wavelength of \(500 \mathrm{~nm}\) or \(510 \mathrm{~nm}\), respectively. What is the energy difference between states \(A\) and \(B ?\)

Step-by-Step Solution

Answer

The energy difference between states \( A \) and \( B \) is approximately \( 0.077 \times 10^{-19} \) Joules.

Step 1: Identify the Formula for Energy

The energy of a photon emitted during a transition is given by the formula \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \, ext{Js}) \), \( c \) is the speed of light \( (3 \times 10^8 \, ext{m/s}) \), and \( \lambda \) is the wavelength of the emitted photon.

Step 2: Calculate Energy of Photon from State A

For the emission from state \( A \), the wavelength is \( 500 \, ext{nm} \). First, convert this to meters: \( 500 \, ext{nm} = 500 \times 10^{-9} \, ext{m} \). Now compute the energy: \[ E_A = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}} \, \text{J} \approx 3.976 \times 10^{-19} \, \text{J} \].

Step 3: Calculate Energy of Photon from State B

For the emission from state \( B \), the wavelength is \( 510 \, ext{nm} \). Convert this to meters: \( 510 \, ext{nm} = 510 \times 10^{-9} \, ext{m} \). Now compute the energy: \[ E_B = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{510 \times 10^{-9}} \, \text{J} \approx 3.899 \times 10^{-19} \, \text{J} \].

Step 4: Calculate the Energy Difference Between States A and B

The energy difference between states \( A \) and \( B \) is the difference in photon energies: \( \Delta E = E_A - E_B \). Calculate it as: \[ \Delta E = 3.976 \times 10^{-19} \, \text{J} - 3.899 \times 10^{-19} \, \text{J} = 0.077 \times 10^{-19} \, \text{J} \].

Key Concepts

Photon Energy

Photon energy is a crucial concept in understanding atomic transitions. It refers to the energy carried by a single photon, the elementary particle of light. In the context of an atom transitioning between energy states, photon energy is emitted as light at a specific wavelength. This wavelength varies depending on the energy difference between the initial and final states of the atom.

The formula to calculate photon energy is given by:

• \( E = \frac{hc}{\lambda} \)

where:

• \( E \) is the energy of the photon.
• \( h \) represents Planck's constant, a fundamental constant in quantum mechanics.
• \( c \) is the speed of light.
• \( \lambda \) is the wavelength of the photon.

This equation helps us determine the energy involved when an atom transitions from a higher energy state to a lower one, emitting a photon in the process.

Accurately calculating photon energy using this formula is essential for comprehending phenomena such as spectral lines, colors of emitted light, and the energy levels of atomic structures.

Planck's Constant

Planck's constant is a pivotal element in the realm of quantum mechanics. Symbolized by \( h \), it has a value of approximately \( 6.626 \times 10^{-34} \) joule-seconds. This constant serves as a bridge between the macroscopic concepts we're familiar with and the microscopic world of atomic and subatomic particles.

Planck's constant appears in several fundamental formulas, including the complete expression for photon energy: \( E = \frac{hc}{\lambda} \). The presence of Planck's constant in this equation signifies the quantized nature of energy. This means energy levels in atoms are discrete and not continuous, which is a core principle of quantum physics.

Understanding this constant is crucial not only for calculating photon energy but also for grasping concepts such as:

• Quantization of energy levels.
• The behavior of electrons within atoms.
• The emission and absorption of electromagnetic radiation.

These ideas are foundational to modern physics and chemistry, shedding light on how matter and energy interact at the smallest scales.

Excited States

In atomic physics, an excited state is a temporary condition where an atom possesses higher energy than its ground state. When an electron in an atom absorbs energy, it can jump to an excited state further from the nucleus.

Excited states are not stable; they exist only for a brief moment before the atom returns to its ground state, releasing the excess energy as a photon. This process of returning to a lower energy state is often referred to as the atom 'relaxing.'

In the given exercise, we consider transitions from two excited states, A and B, to the ground state, where each state emits a photon at slightly different wavelengths (500 nm and 510 nm, respectively). These different wavelengths reveal that the energy levels of these excited states differ slightly, which can be quantified using the photon energy equation.

Understanding excited states is vital for studying:

• The emission spectra of elements.
• The mechanisms of lasers and fluorescent lights.
• Chemical reactions where energy transfer occurs.

This knowledge allows scientists to analyze how energy transitions affect atomic and molecular behavior, leading to innovations across various scientific disciplines.