Problem 3 - University Physics with Modern Physics

Question

A 75-W light source consumes 75 W of electrical power. Assume all this energy goes into emitted light of wavelength 600 nm. (a) Calculate the frequency of the emitted light. (b) How many photons per second does the source emit? (c) Are the answers to parts (a) and (b) the same? Is the frequency of the light the same thing as the number of photons emitted per second? Explain.

Step-by-Step Solution

Answer

Frequency is about \(5 \times 10^{14}\) Hz; emits about \(2.26 \times 10^{20}\) photons/sec. Frequency is not the same as the number of photons emitted.

Step 1: Understanding Wavelength and Frequency

The wavelength of the light is given as 600 nm. To calculate frequency, we use the formula \( c = \lambda u \), where \( c \) is the speed of light (\( 3 \times 10^8 \) m/s), \( \lambda \) is the wavelength, and \( u \) is the frequency. Convert 600 nm to meters: \( 600 \times 10^{-9} \) m.

Step 2: Calculate Frequency

Rearrange the formula \( \lambda u = c \) to solve for \( u \): \( u = \frac{c}{\lambda} \). Substitute the values to find frequency; \( u = \frac{3 \times 10^8}{600 \times 10^{-9}} \approx 5 \times 10^{14} \) Hz.

Step 3: Calculate Energy of One Photon

The energy \( E \) of one photon can be calculated using the formula \( E = h u \), where \( h \) is Planck's constant (\( 6.63 \times 10^{-34} \) Js). Thus, \( E = 6.63 \times 10^{-34} \times 5 \times 10^{14} \approx 3.315 \times 10^{-19} \) J/photon.

Step 4: Calculate Number of Photons Emitted

The power of the light source is 75 W, which means 75 J/s are being emitted. To find the number of photons emitted per second, divide power by the energy per photon: \( \frac{75}{3.315 \times 10^{-19}} \approx 2.26 \times 10^{20} \) photons per second.

Key Concepts

Frequency Calculation

Frequency is a way to describe how often a wave oscillates in a given amount of time. Specifically for light, frequency is how many wave crests pass by a point each second. It's a central property of electromagnetic waves and is measured in hertz (Hz), where 1 Hz equals one wave per second.

To calculate the frequency of light when you know its wavelength, you use the equation \( c = \lambda u \), where \( c \) is the speed of light (\(3 \times 10^8\) meters per second), \( \lambda \) is the wavelength, and \( u \) is the frequency. When you rearrange this formula to solve for frequency, it becomes \( u = \frac{c}{\lambda} \).

For example, if a light source emits a wavelength of 600 nm (nanometers), you first convert this wavelength into meters: \(600 \times 10^{-9}\) m. Plugging this into the frequency formula, you divide the speed of light by the wavelength in meters: \( u = \frac{3 \times 10^8}{600 \times 10^{-9}} \). This calculation results in a frequency of approximately \(5 \times 10^{14}\) Hz.

This frequency tells you how rapidly the light waves are oscillating, which is vital for differentiating colors in the visible spectrum.

Photon Energy

When it comes to photons, they are the tiny packets of energy that constitute light. Each photon carries energy, and the amount of this energy depends on the light's frequency.

To calculate the energy of a single photon, we use Planck's equation: \( E = h u \). Here, \( E \) represents the energy, \( h \) is Planck's constant (\(6.63 \times 10^{-34}\) Joule seconds), and \( u \) is the frequency of the light. The energy of a photon is directly proportional to its frequency: the higher the frequency, the more energy a photon possesses.

Applying this to our example with light of frequency \(5 \times 10^{14}\) Hz, the energy of one photon is \( E = 6.63 \times 10^{-34} \times 5 \times 10^{14} \approx 3.315 \times 10^{-19} \) Joules. This illustrates how each photon, although extremely small, carries a specific quantum of energy, which determines its interactions with matter.

Wavelength and Frequency Relationship

The relationship between wavelength and frequency of light is a fundamental aspect of wave physics. Wavelength and frequency are inversely related, meaning when one increases, the other decreases. This relationship is expressed by the formula \( c = \lambda u \), signifying the product of wavelength \( \lambda \) and frequency \( u \) equals the constant speed of light \( c \).

This explains why light with a longer wavelength has a lower frequency, and conversely, light with a shorter wavelength has a higher frequency. This interplay is crucial because it dictates the position of light within the electromagnetic spectrum, from gamma rays with very high frequencies and short wavelengths to radio waves with low frequencies and long wavelengths.

Understanding this relationship helps in identifying different types of electromagnetic waves based on their properties, such as visible light which falls between approximately 400 nm to 700 nm in wavelength, corresponding to frequencies ranging from about \(4.3 \times 10^{14}\) Hz to \(7.5 \times 10^{14}\) Hz.