Problem 5 - College Physics

Question

The predominant wavelength emitted by an ultraviolet lamp is \(248 \mathrm{nm} .\) If the total power emitted at this wavelength is \(12.0 \mathrm{~W}\), how many photons are emitted per second?

Step-by-Step Solution

Answer

Approximately \(1.50 \times 10^{19}\) photons are emitted per second.

Step 1: Understand the Problem and Gather Information

We need to calculate the number of photons emitted per second by the UV lamp, given its power output and wavelength. The lamp emits a power of 12.0 W at a wavelength of 248 nm.

Step 2: Calculate the Energy of a Single Photon

The energy of a single photon is given by the formula \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \) J·s), \( c \) is the speed of light (\( 3.00 \times 10^8 \) m/s), and \( \lambda \) is the wavelength in meters. First, convert the wavelength from nm to meters: \( 248 \text{ nm} = 248 \times 10^{-9} \text{ m} \). Then, calculate the energy:\[E = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{248 \times 10^{-9}} \approx 8.02 \times 10^{-19} \text{ J}\]

Step 3: Calculate the Number of Photons Emitted per Second

Using the total power output and the energy of one photon, the number of photons emitted per second \( n \) can be calculated using the formula \( n = \frac{P}{E} \), where \( P \) is the power in watts (joules per second). Substitute the known values:\[n = \frac{12.0}{8.02 \times 10^{-19}} \approx 1.50 \times 10^{19} \text{ photons per second}\]

Step 4: Double-check the Calculations

Re-calculate each step and verify the unit conversions and final calculation for accuracy to ensure reliable results.

Key Concepts

Ultraviolet Light

Ultraviolet (UV) light is a type of electromagnetic radiation. It exists beyond the violet end of the visible light spectrum. While we cannot see UV light, it has significant effects in our world.

• UV wavelengths range from about 10 nm to 400 nm.
• UV light is classified into three types: UVA, UVB, and UVC based on their wavelengths.
• It is commonly used in applications such as sterilization and fluorescence.

In this exercise, the UV lamp emits light predominantly at a wavelength of 248 nm. This places it in the UVC category, which is the most energetic and has a range from about 100 to 280 nm. UVC is often used for disinfecting because it can destroy bacteria and viruses by causing changes in their DNA.
Understanding UV light, particularly the types and their properties, is essential to grasping how it interacts with materials and its practical applications.

Planck's Constant

Planck's constant is a fundamental constant in physics, denoted by the symbol \( h \). It plays a crucial role in the field of quantum mechanics.

• Its value is approximately \( 6.626 \times 10^{-34} \text{ J·s} \).
• It relates the energy of a photon to its frequency: \( E = hf \).
• The constant helps explain phenomena like photon emission and the quantization of energy levels in atoms.

In the context of this exercise, Planck's constant is part of the calculation for determining the energy of a single photon emitted at a specific wavelength. Understanding this constant is essential because it bridges the gap between classical and quantum physics, helping describe the behavior of light and particles on an atomic scale.

Wavelength Conversion

Wavelength conversion is essential in scientific calculations when the initial measurement is not in the standard unit. Wavelength is typically measured in nanometers (nm) in the context of light. Nonetheless, for equations requiring consistency in units, such as meters in energy calculations, conversion becomes necessary.

• To convert from nanometers to meters, use the conversion: \(1 \text{ nm} = 1 \times 10^{-9} \text{ m}\).
• In this exercise, the UV wavelength of 248 nm converts to meters: \(248 \times 10^{-9} \text{ m}\).

Accurate wavelength conversion is vital to performing precise calculations, ensuring that all components of an equation are compatible, thereby leading to reliable and correct results.

Energy of Photons

The energy of photons is a critical subject in physics, particularly in quantum mechanics and electromagnetic theory. Understanding how to calculate the energy of a single photon is vital for numerous scientific fields.

• The formula to find a photon's energy is \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant, \( c \) is the speed of light (\(3.00 \times 10^8 \text{ m/s}\)), and \( \lambda \) is the wavelength in meters.
• In this context, the energy for a photon with a wavelength of 248 nm is calculated as approximately \( 8.02 \times 10^{-19} \text{ J}\).
• Photon energy is linked to frequency and wavelength; a shorter wavelength corresponds to higher energy.

Calculating photon energy helps us understand phenomena like power distribution in light sources, photoelectric effects, and the behaviors of materials under illumination. It is a foundational concept for interpreting and applying the principles of physics to real-world applications.