Problem 26 - University Physics with Modern Physics

Question

(a) If the average frequency emitted by a 120-W light bulb is 5.00 \(\times\) 10\(^{14}\) 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 cm\(^2\) per second if the light is emitted uniformly in all directions?

Step-by-Step Solution

Answer

(a) 3.621 × 10¹⁹ photons/second; (b) Distance: ~85.05 meters.

Step 1: Calculate Power Emitted as Visible Light

First, identify the power that is actually emitted as visible light. Since 10.0\(\%\) of the input power is emitted as visible light, multiply the total power by this percentage:\[P_{\text{visible}} = 120 \text{ W} \times 0.10 = 12 \text{ W}\]

Step 2: Determine Energy Per Photon

The energy of a photon can be calculated using the formula: \[E = h \times f\]where \(h\) is Planck's constant \((6.626 \times 10^{-34} \text{ J} \cdot \text{s})\) and \(f\) is the frequency of the light. Given that the frequency is \(5.00 \times 10^{14} \text{ Hz}\), the energy per photon is:\[E = 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \times 5.00 \times 10^{14} \text{ Hz} = 3.313 \times 10^{-19} \text{ J}\]

Step 3: Calculate Number of Photons Emitted Per Second

Now, we can find out how many photons are emitted per second by dividing the total power emitted as visible light by the energy per photon:\[\text{Number of photons per second} = \frac{12 \text{ W}}{3.313 \times 10^{-19} \text{ J/photon}}\]This calculation gives us approximately:\[3.621 \times 10^{19} \text{ photons/second}\]

Step 4: Calculate Total Emission Rate

We now know that the bulb emits \(3.621 \times 10^{19}\) photons per second uniformly in all directions.

Step 5: Find Area and Distance for Given Photon Density

To find the distance where the emission corresponds to \(1.00 \times 10^{11}\) photons per cm\(^2\) per second, determine the area that receives the photons. For a sphere with radius \(r\):\[A = 4\pi r^2\]Set up the equation for the photon density: \[1.00 \times 10^{11} \text{ photons/cm}^2 \text{ converted to m}^2 = 1.00 \times 10^{15} \text{ photons/m}^2\]Now solve for \(r\):\[3.621 \times 10^{19} \text{ photons/second} = 1.00 \times 10^{15} \text{ photons/m}^2 \times 4 \pi r^2\]Solve for \(r^2\):\[r^2 = \frac{3.621 \times 10^{19}}{1.00 \times 10^{15} \times 4\pi}\]Thus, calculate \(r\):\[r \approx 85.05 \text{ meters}\]

Step 6: Conclude with Final Distances

We found that at a distance of approximately 85.05 meters, the bulb emits photons at the specified density of \(1.00 \times 10^{11}\) photons per cm\(^2\) per second.

Key Concepts

Energy of a Photon

To understand the energy of a photon, it's important to know that photons are basic units of light. Their energy is given by the equation \[ E = h \times f \],where \( E \) stands for energy, \( h \) is Planck's constant, and \( f \) is the frequency of the photons.

• The frequency \( f \) refers to how many wave peaks pass a certain point per second, measured in Hertz (Hz).
• This relationship means that higher frequency photons have more energy.

By knowing the frequency and using Planck's constant, we can determine exactly how much energy a single photon of light carries. In our exercise, with a frequency of \( 5.00 \times 10^{14} \) Hz and Planck's constant \( 6.626 \times 10^{-34} \) J\cdot s, the photon energy is calculated. This particular calculation tells us that each photon has quite a small amount of energy, showing why large numbers of photons are necessary to make visible light.

Planck's Constant

Planck's constant \(h\) is a fundamental quantity in quantum mechanics, expressed as \(6.626 \times 10^{-34} \text{ J} \cdot \text{s}\). It's named after Max Planck, a scientist who was key to developing quantum theory.

• Planck's constant links the energy and frequency of a photon, working as a bridge between the wave and particle nature of light.
• Because of its very small value, it demonstrates why the quantum effects are typically significant only at atomic and subatomic levels.

Planck's constant plays a crucial role in calculations involving the energy of photons. In everyday terms, this means it helps us understand how much energy each tiny packet of light (photon) carries when we know its frequency.

Visible Light Power Calculation

To find out how much power in visible light comes from a light source—like the bulb in our problem—you first identify the percentage of the bulb's total power that is emitted as visible light.

• Only part of the total power is actually converted into visible light; the rest can become heat or invisible radiation.
• In this example, only 10% of the input power of 120 watts was used for visible light: \(12 \text{ W} \).

After establishing the specific power for visible light, one can determine the number of photons emitted. This involves dividing the visible light's power by the energy per photon. Thus, knowing both the energy of each photon and the total visible light power allows for back-of-the-envelope calculations about how brightly the bulb shines in terms of photon counts per second.

Photon Density

Photon density describes the concentration of photons over a specific area. It is expressed as the number of photons per unit area, such as per square centimeter or square meter.

• A higher photon density implies a brighter perception of light.
• In our exercise, photon density is crucial to finding the distance at which a defined number of photons per square centimeter is received from the bulb.

Converting the required density to a consistent unit (from cm\(^{2}\) to m\(^{2}\)) is often necessary for calculations. The concept of photon density helps in understanding how photon emissions spread over distances and affect what we perceive as brightness at different spots in relation to a light source.

Uniform Emission

Uniform emission means that the light from a source, like a bulb, spreads equally in all directions. This is crucial when calculating how light behaves over distances.

• If the light were not uniform, some areas would get more photons and some less, affecting brightness perception.
• For a sphere surrounding the light source, the light's spread over a surface area of \(4\pi r^2\) (where \( r \) is the radius).

In calculations involving uniform light sources, you assume that every part of the surface area at a given distance receives the same number of photons. Understanding uniform emission helps predict light intensity where it is emitted evenly, offering a baseline for various optics studies and practical applications.