Problem 92 - General Chemistry: Principles and Modern Applications

Question

Infrared lamps are used in cafeterias to keep food warm. How many photons per second are produced by an infrared lamp that consumes energy at the rate of \(95 \mathrm{W}\) and is \(14 \%\) efficient in converting this energy to infrared radiation? Assume that the radiation has a wavelength of \(1525 \mathrm{nm}\).

Step-by-Step Solution

Answer

The infrared lamp produces \(3.24 x 10^{20}\) photons per second.

Step 1: Calculate the Energy per Photon

Use Planck's formula to calculate the energy of a single photon: \(E = \frac{hc}{λ}\), where \(h\) is Planck's constant \((6.626 x 10^{-34} Js)\), \(c\) is the speed of light \((3 x 10^8 ms^{-1})\), and \(λ\) is the wavelength which was given as 1525 nm or \(1.525 x 10^{-6} m\). When these values are substituted in, the energy per photon is found to be \(4.10 x 10^{-19} Joules\).

Step 2: Calculate Total Energy Rougly Converted to Photons

Given the lamp's power consumption and efficiency, find the energy being converted to photons per second. The power consumption is 95 W (which means it uses 95 Joules of energy per second), and efficiency is 14%, which should be represented as 0.14 when calculating. Multiply these together: \(95 W x 0.14 = 13.3 W\), which means about 13.3 Joules of energy are converted into photons every second.

Step 3: Calculate the Number of Photons per Second

Using the fact that each photon carries energy \(4.10 x 10^{-19} Joules\), and the lamp produces 13.3 Joules of photon energy per second, find out how many photons this corresponds to by dividing the total energy per second by the energy per photon: \(13.3J / (4.10 x 10^{-19}J/photon) = 3.24 x 10^{20} photons/second\). Thus the infrared lamp produces this amount of photons every second.

Key Concepts

Planck's Constant

Planck's constant, denoted by the symbol \( h \), is a fundamental constant in physics that contributes significantly to the quantum mechanics world. It's a small number: \( 6.626 \times 10^{-34} \) Joule-seconds. This value may seem minuscule, but it plays a crucial role in understanding how energy is quantized. When dealing with photons or "particles of light," Planck's constant is part of the formula that helps determine the energy those photons carry. By knowing the wavelength of the light or radiation, and indeed the speed of light, we can use Planck's constant in the formula \( E = \frac{hc}{\lambda} \) to calculate the energy contained in one photon. This relationship highlights how radiation at different wavelengths translates into different energies. Planck's constant connects these seemingly disparate ideas with a single, unifying measurement.

Photon Energy Calculation

Calculating the energy of a single photon involves using a formula that combines Planck’s constant, the speed of light \( c \), and the wavelength \( \lambda \) of the photon. This formula is \( E = \frac{hc}{\lambda} \).

• Use Planck’s constant (\( h \)): This is always \( 6.626 \times 10^{-34} \) J\cdot s.
• Speed of light (\( c \)): Use \( 3 \times 10^8 \) m/s for all calculations.
• Wavelength (\( λ \)): This must be in meters.

For instance, in an exercise where the wavelength is 1525 nm, convert this to meters: \( 1.525 \times 10^{-6} \) m. Substituting these values gives us the photon's energy: \( 4.10 \times 10^{-19} \) Joules. Such calculations let us compare energies of photons from different parts of the electromagnetic spectrum, providing insights into their potential application or impact.

Efficiency of Energy Conversion

The concept of efficiency in energy conversion is pivotal in understanding how effectively a device translates input energy to desired output energy. In the context of infrared radiation, the efficiency measures how well a lamp converts electrical energy into infrared light.

• Formula for Efficiency: Efficiency \( = \frac{\text{Useful energy output}}{\text{Total energy input}} \times 100 \% \)
• Example with Infrared Lamp: If the lamp consumes 95 watts of power but only 14% is converted to infrared, then the effective energy conversion is \( 95 \text{ W} \times 0.14 = 13.3 \text{ W} \).

This means 13.3 watts are used for heating through infrared radiation. Understanding efficiency helps in improving device performance, saving energy, and using resources wisely, which is essential for technological and environmental advancements.