Question

Only a fraction of the electrical energy supplied to a tungsten lightbulb is converted to visible light. The rest of the energy shows up as infrared radiation (that is, heat). A 75-W lightbulb converts 15.0 percent of the energy supplied to it into visible light (assume the wavelength to be \(550 \mathrm{nm}\) ). How many photons are emitted by the lightbulb per second? \((1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s} .)\)

Short answer

The number of photons emitted by the lightbulb per second is approximately \(3.11 \times 10^{19}\).

Step-by-step solution

Calculate the power converted to light

The total power of the lightbulb is 75W but only a fraction of this power is converted into visible light. The fraction converted into visible light is given as 15%, so you must calculate how much actual power that represents by multiplying 75W by 0.15. That will be \(75W \times 0.15 = 11.25W\).

Find the energy of a single photon

The energy of each photon (\(E\)) is found using the equation, \(E = \frac{hc}{\lambda}\), where \(h\) is Planck’s constant (\(h = 6.626 \times 10^{-34} Js\)), \(c\) is the speed of light (\(c = 3.00 \times 10^{8} m/s\)) and \(\lambda\) represents the wavelength (\(\lambda = 550 \times 10^{-9} m\)). Substituting these values into the equation gives, \(E = \frac{6.626 \times 10^{-34} Js \times 3.00 \times 10^{8} m/s}{550 \times 10^{-9} m} = 3.62 \times 10^{-19} J\).

Find the number of photons

Power is defined as the amount of work done or energy converted per unit of time. In this context, it represents the amount of energy emitted per second. To find out how many photons are being emitted by the bulb per second, therefore, the energy being emitted as light per second (or the power that is enitted as light term calculated in step 1) is divided by the energy of one photon (calculated in step 2). This gives you: \(\frac{11.25 J}{3.62 \times 10^{-19} J/photon} = 3.11 \times 10^{19} photons/s\).

Key concepts

Energy Conversion in Lightbulbs

Lightbulbs are common sources of artificial light and working examples of energy conversion. Specifically, they transform electrical energy into electromagnetic radiation, which includes both visible light and other forms of energy like infrared radiation.

A typical incandescent lightbulb, like one rated at 75 watts, uses a tungsten filament. When electricity passes through the filament, it heats up due to its resistance to the electric current. This heat causes the filament to glow, emitting light. However, not all energy is converted to light; a significant portion is lost as heat.

In our example, it's given that only 15% of the electrical energy supplied to a 75-W lightbulb is converted into visible light. To understand the energy conversion efficiency, you can calculate the energy used as visible light by multiplying the power rating of the lightbulb by the percentage of energy converted to light. Hence, a 75-W bulb with 15% efficiency emits 11.25 W of visible light.

Planck's Constant Application

Planck's constant is a fundamental quantity in physics named after the physicist Max Planck. It plays a vital role in the quantum theory of energy and light. The constant, represented by the symbol 'h,' connects the amount of energy a photon carries with the frequency of its electromagnetic wave.

The formula to calculate the energy of a single photon is given by the equation:
\[E = \frac{hc}{\lambda}\]
where \(E\) is the energy of the photon, \(h\) is Planck's constant (approximately \(6.626 \times 10^{-34} J\cdot s\)), \(c\) is the speed of light (approximately \(3.00 \times 10^8 m/s\)), and \(\lambda\) is the wavelength of the light.

By applying Planck's constant in our calculation, we can determine the energy of a photon with a wavelength of 550 nm, leading us to understand the photon's contribution to the light emitted by a bulb.

Speed of Light in Equations

The speed of light, typically denoted as \(c\), is a crucial constant in physics, with a value approximately equal to \(3.00 \times 10^{8} m/s\). It signifies how fast light travels in a vacuum and is essential in many equations, not only in physics but also in engineering and technology.

In the context of photon emission calculations, the speed of light is used together with Planck’s constant to determine the energy of a single photon. As you can see in the provided solution, integrating the speed of light in the equation \(E = \frac{hc}{\lambda}\), allows us to calculate the energy for each photon emitted depending on its wavelength.

Therefore, understanding how to use this constant is fundamental for anyone studying the properties and behavior of light, as well as its interaction with matter.

Wavelength of Visible Light

The wavelength of light is a measure of the distance between identical points in the adjacent cycles of a waveform signal propagated in space. In terms of visible light, it determines the color we perceive. The human eye can typically see wavelengths from about 380 to 750 nanometers (nm). Each color within this spectrum has a unique wavelength, with violet on the short wavelength end and red on the long-wavelength end.

The problem provided deals with light of a 550 nm wavelength, placing it within the green part of the visible spectrum. When it comes to the emission of photons by a lightbulb, different wavelengths correspond to photons of different energies.

The energy of a photon is inversely proportional to its wavelength, as evidenced by the equation \(E = \frac{hc}{\lambda}\). Therefore, photons of visible light with shorter wavelengths carry more energy than those with longer wavelengths. In practical applications, this influences the color and intensity of the light that lightbulbs emit.