Problem 11 - Chemistry and Chemical Reactivity

Question

An energy of \(3.3 \times 10^{-19} \mathrm{J} /\) atom is required to cause a cesium atom on a metal surface to lose an electron. Calculate the longest possible wavelength of light that can ionize a cesium atom. In what region of the electromagnetic spectrum is this radiation found?

Step-by-Step Solution

Answer

602 nm light can ionize cesium, found in the visible spectrum.

Step 1: Understand the Relationship Between Energy and Wavelength

The energy of a photon is related to its wavelength by the formula: \( E = \frac{hc}{\lambda} \), where \( E \) is energy, \( h \) is Planck's constant \( (6.626 \times 10^{-34} \ \text{J} \cdot \text{s}) \), \( c \) is the speed of light \( (3 \times 10^8 \ \text{m/s}) \), and \( \lambda \) is the wavelength.

Step 2: Rearrange the Formula for Wavelength

To find the wavelength, rearrange the formula \( E = \frac{hc}{\lambda} \) to solve for \( \lambda \):\[ \lambda = \frac{hc}{E} \]

Step 3: Insert Known Values into the Formula

Substitute \( h = 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \), \( c = 3 \times 10^8 \text{ m/s} \), and \( E = 3.3 \times 10^{-19} \text{ J} \) into the equation:\[ \lambda = \frac{(6.626 \times 10^{-34} \text{ J} \cdot \text{s})(3 \times 10^8 \text{ m/s})}{3.3 \times 10^{-19} \text{ J}} \]

Step 4: Perform the Calculation

Calculate the value:\[ \lambda = \frac{1.9878 \times 10^{-25} \text{ J} \cdot \text{m}}{3.3 \times 10^{-19} \text{ J}} = 6.02 \times 10^{-7} \text{ m} \]This is equivalent to \( 602 \text{ nm} \) (since \( 1 \text{ nm} = 10^{-9} \text{ m} \)).

Step 5: Determine the Region of the Electromagnetic Spectrum

The wavelength \( 602 \text{ nm} \) falls in the visible spectrum, which ranges from approximately 400 nm to 700 nm.

Key Concepts

Planck's constant

Planck's constant is a fundamental constant in physics with a value of approximately \(6.626 \times 10^{-34} \ \text{J} \cdot \text{s}\). It was introduced by Max Planck when he was solving the black-body radiation problem at the beginning of the 20th century. This constant describes the relationship between the energy of a photon and the frequency of its corresponding electromagnetic wave. The energy \(E\) of a photon can be determined using the formula:\[E = h u\]where \(u\) is the frequency and \(h\) is Planck's constant. This relationship demonstrates that energy is quantized in discrete units called quanta. Understanding Planck's constant helps us grasp how energy levels in atoms are determined, which is crucial for explaining phenomena like the photoelectric effect, where electrons are emitted from materials when they absorb light energy.

Photon energy

Photon energy is a key concept in understanding how light interacts with matter. A photon is an elementary particle, the quantum of the electromagnetic field including electromagnetic radiation such as light.The energy of a photon is directly proportional to its frequency and inversely proportional to its wavelength. The relationship between these quantities is defined by the equation:\[ E = \frac{hc}{\lambda} \]where:

• \(E\) is the photon energy,
• \(h\) is Planck's constant \((6.626 \times 10^{-34} \ \text{J} \cdot \text{s})\),
• \(c\) is the speed of light \((3 \times 10^8 \ \text{m/s})\), and
• \(\lambda\) is the wavelength.

This equation allows you to calculate the energy needed to ionize an atom or to determine the type of electromagnetic radiation based on its energy. Photon energy is an essential factor in various applications, from explaining the colors in visible light to the mechanisms of solar panels and lasers.

Electromagnetic spectrum

The electromagnetic spectrum encompasses all types of radiation that travel at the speed of light, organized by their wavelengths and frequencies. It ranges from long-wavelength, low-frequency radio waves to short-wavelength, high-frequency gamma rays. The visible spectrum, which is part of this electromagnetic spectrum, includes wavelengths from approximately 400 to 700 nanometers. Each color of visible light corresponds to a specific range of wavelengths:

• Violet: 400-450 nm
• Blue: 450-495 nm
• Green: 495-570 nm
• Yellow: 570-590 nm
• Orange: 590-620 nm
• Red: 620-700 nm

Beyond the visible spectrum, there are infrared, ultraviolet, X-rays, and others, each useful for different technological and scientific purposes. Understanding the electromagnetic spectrum is crucial in various fields such as astronomy, communications, and medical imaging — further illustrating the broad impact of electromagnetic radiation on technology and daily life.