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Chapter 38: Problem 30

(a) If the average frequency emitted by a 200 -W light bulb is \(5.00 \times 10^{14} \mathrm{Hz}\) , and 10.0\(\%\) of the input power is emitted as visible light, approximately how many visible-light photons are emitted per second? (b) At what distance would this correspond to \(1.00 \times 10^{11}\) visible-light photons per square centimeter per second if the light is emitted uniformly in all directions?

Short Answer

Expert verified
The light bulb emits approximately \( 3.01 \times 10^{20} \) photons per second. The corresponding distance is about 0.87 meters.

Step by step solution

01

Calculate Power Emitted as Visible Light

First, we note that the light bulb has a total power of 200 W. Given that 10% of this power is emitted as visible light, we can calculate the power emitted as visible light using the formula: \[ P_{visible} = 0.10 \times 200 \, \text{W} \]Compute this value to find out how much power is emitted as visible light.
02

Determine the Energy of a Single Photon

To find the energy of a single photon, we use the formula: \[ E = h \cdot f \]where \( h \) is Planck's constant \( 6.626 \times 10^{-34} \, \text{Js} \) and \( f \) is the frequency \( 5.00 \times 10^{14} \, \text{Hz} \). Calculate the energy \( E \) of one photon based on this frequency.
03

Calculate Number of Photons Emitted per Second

Using the power of visible light \( P_{visible} \) calculated in Step 1 and the energy per photon determined in Step 2, we can find the number of photons emitted per second by rearranging the formula: \[ n = \frac{P_{visible}}{E} \]where \( n \) is the number of photons per second.
04

Determine the Distance Corresponding to a Photon Intensity

We want to find the distance at which the photon intensity is \( 1.00 \times 10^{11} \) visible-light photons per square centimeter per second. First, convert this intensity to photons per square meter using the conversion factor \( 1 \, \text{cm}^2 = 10^{-4} \, \text{m}^2 \).Then use the formula for intensity distributed over a sphere: \[ I = \frac{n}{4 \pi r^2} \]where \( n \) is the total number of photons emitted per second and \( r \) is the distance from the light source. Rearrange to solve for \( r \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Light Bulb Power
Understanding light bulb power is essential when tackling problems that involve energy emissions. In this exercise, we start with a 200-Watt light bulb. This indicates that the bulb uses 200 Joules of energy every second. However, not all this energy gets converted into visible light.
Typically, only a small portion of this energy appears as visible light. Here, 10% of the total power is what we need to consider. So, the calculation involves taking 10% of 200 Watts, which is 20 Watts emitted as visible light. This value signifies how much energy is used every second solely for visible light emission.
Visible Light Emission
Visible light emission refers to the portion of light that is within the range our eyes can detect, generally between wavelengths of 380 nm to 750 nm. In our scenario, the light bulb emits its visible light at a frequency of \(5.00 \times 10^{14} \text{Hz}\).
It's significant because this frequency falls within the range of visible light. Therefore, among all the energy used by the bulb, we focus on this small segment that humans perceive as light. The frequency gives us insight into the specific light color, which in everyday terms, can range from red to violet within the visible spectrum.
Photon Energy Calculation
Photon energy is the energy carried by a single photon and is crucial for understanding how many photons are emitted. This energy is calculated using Planck's formula: \( E = h \cdot f \) where \(h\) is Planck's constant (\(6.626 \times 10^{-34} \text{Js}\)) and \(f\) is the frequency (\(5.00 \times 10^{14} \text{Hz}\)).
By plugging in these values, you'll calculate the energy of one photon of visible light. This small energy value becomes crucial in the next steps, where it helps determine how many such photons are emitted with the total visible power available.
Photon Intensity
Photon intensity measures how many photons hit a specific area over time, often defined as photons per square centimeter per second. This exercise calculates the distance at which a photon intensity of \(1.00 \times 10^{11}\) photons/cm虏/s is achieved.
The light bulb emits photons uniformly in all directions, conceptualized as emitting over a sphere's surface. Thus, the intensity is distributed over that sphere. We're using \(I = \frac{n}{4 \pi r^2}\) where \(n\) corresponds to the number of photons emitted every second. By manipulating this formula, you can solve for the distance \(r\), which achieves the desired intensity.
Planck's Constant
Planck's constant (\(h\)) is a fundamental ingredient in quantum mechanics, representing the proportionality factor between energy and frequency. In this problem, it is key to calculating the energy of each photon emitted by the light bulb.
It has a fixed value of \(6.626 \times 10^{-34} \text{Js}\). When you multiply it by the frequency of the visible light, you get the photon energy. This constant ties energy and frequency together and reflects the quantum nature of light, explaining why increasing the frequency of light raises its energy.

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Most popular questions from this chapter

Cp Blo Removing Vascular Lesions. A pulsed dye laser emits light of wavelength 585 nm in \(450-\mu\) s pulses. Because this wavelength is strongly absorbed by the hemoglobin in the blood, the method is especially effective for removing various types of blemishes due to blood, such as port-wine-colored birthmark. To get a reasonable estimate of the power required for such laser surgery, we can model the blood as having the same specific heat and heat of vaporization as water \(\left(4190 \mathrm{J} / \mathrm{kg} \cdot \mathrm{K}, 256 \times 10^{6} \mathrm{J} / \mathrm{kg}\right) .\) Suppose that each pulse must remove 2.0\(\mu \mathrm{g}\) of blood by evaporating it, starting at \(33^{\circ} \mathrm{C}\) . (a) How much energy must each pulse deliver to the blemish? (b) What must be the power output of this laser? (c) How many photons does each pulse deliver to the blemish?

A photon with wavelength 0.1100 nm collides with a free electron that is initially at rest. After the collision the wavelength is 0.1132 \(\mathrm{nm}\) (a) What is the kinetic energy of the electron after the collision? What is its speed? (b) If the electron is suddenly stopped (for example, in a solid target), all of its kinetic energy is used to create a photon. What is the wavelength of this photon?

(a) Derive an expression for the total shift in photon wavelength after two successive Compton scatterings from electrons at rest. The photon is scattered by an angle \(\theta_{1}\) in the first scattering and by \(\theta_{2}\) in the second. (b) In general, is the total shift in wavelength produced by two successive scatterings of an angle \(\theta / 2\) the same as by a single scattering of \(\theta\) ? If not, are there any specific values of \(\theta\) , other than \(\theta=0^{\circ}\) , for which the total shifts are the same? (c) Use the result of part (a) to calculate the total wavelength shift produced by two successive Compton scatterings of \(30.0^{\circ}\) each. Express your answer in terms of \(h / m c .\) (d) What is the wavelength shift produced by a single Compton scattering of \(60.0^{\circ}\) . Compare to the answer in part (c).

A photon of green light has a wavelength of 520 \(\mathrm{nm} .\) Find the photon's frequency, magnitude of momentum, and energy. Express the energy in both joules and electron volts.

A 75 -W light source consumes 75 W of electrical power. Assume all this energy goes into emitted light of wavelength 600 \(\mathrm{nm}\) . (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.
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