Question

Electric power is measured in watts \((1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}) .\) About \(95 \%\) of the power output of an incandescent bulb is converted to heat and \(5 \%\) to light. If \(10 \%\) of that light shines on your chemistry textbook, how many photons per second shine on the book from a 75-W bulb? (Assume that the photons have a wavelength of \(550 \mathrm{nm} .\) )

Short answer

1.038 × 10¹⁸ photons/second.

Step-by-step solution

Calculate the Power Converted to Light

The total power output of the bulb is 75 W. Since 5% of this power is converted to light, the power output in terms of light is \[P_{light} = 75 \text{ W} \times 0.05 = 3.75 \text{ W}.\]

Determine the Power Shining on the Textbook

Only 10% of the light shines on the textbook. \[P_{textbook} = 3.75 \text{ W} \times 0.10 = 0.375 \text{ W}.\]

Convert the Power to Energy per Second in Joules

Since 1 W (watt) is equivalent to 1 J/s (joule per second), \[ Energy_{textbook} = 0.375 \text{ W} = 0.375 \text{ J/s}.\]

Calculate the Energy of a Single Photon

Use the formula for the energy of a photon: \[E_{photon} = \frac{hc}{\lambda},\] where \(h\) is Planck's constant \(6.626 \times 10^{-34} \text{ J}\text{s}\), \(c\) is the speed of light \(3 \times 10^8 \text{ m/s}\), and \(\lambda\) is the wavelength \(550 \times 10^{-9} \text{ m}\). Plug in the values: \[ E_{photon} = \frac{(6.626 \times 10^{-34} \text{ J}\text{s})(3 \times 10^8 \text{ m/s})}{550 \times 10^{-9} \text{ m}} = 3.615 \times 10^{-19} \text{ J/Photon}.\]

Calculate the Number of Photons per Second

Divide the energy per second shining on the textbook by the energy of a single photon: \[ \text{Photons/second} = \frac{0.375 \text{ J/s}}{3.615 \times 10^{-19} \text{ J/Photon}} = 1.038 \times 10^{18} \text{ Photons/second}.\]

Key concepts

Power Conversion

Power is a measure of how much energy is used or produced per unit time. In this exercise, the power output of our light bulb is 75 watts (W). Because watts are equivalent to joules per second (J/s), this means the bulb uses or produces 75 joules of energy every second.
Not all the energy from the light bulb becomes visible light. In fact, only 5% of the 75 W output is converted to light while the rest turns into heat. Calculating this gives us:
\[ 75 \text{ W} \times 0.05 = 3.75 \text{ W}. \]
This means 3.75 joules of energy per second are actually emitted as light.

Energy of Photons

Photons are tiny packets of light energy. The energy of a single photon can be calculated using the formula:
\[ E_{photon} = \frac{hc}{\lambda}, \]
where \h is Planck's constant (\(6.626 \times 10^{-34} \text{ J}\text{s})\), \c is the speed of light (\(3 \times 10^8 \text{ m/s})\), and \lambda is the wavelength of the light (\(550 \times 10^{-9} \text{ m})\).
This equation arises from the wave-particle duality of light. When you plug in the given values:
\[ E_{photon} = \frac{ \left( 6.626 \times 10^{-34} \text{ J}\text{s} \right) \left( 3 \times 10^8 \text{ m/s} \right) }{ 550 \times 10^{-9} \text{ m} } = 3.615 \times 10^{-19} \text{ J}. \]
So each photon carries an energy of 3.615 x 10^-19 joules.

Planck's Constant

Planck's constant (\(6.626 \times 10^{-34} \text{ J}\text{s}\)) is a fundamental constant in physics. It plays a crucial role in the field of quantum mechanics by relating the energy of a photon to the frequency of its corresponding electromagnetic wave. Essentially, it acts as a bridge between the wave and particle nature of light.
In this exercise, we use Planck’s constant to determine the energy of photons emitted by the bulb. This constant helps make sense of how much energy individual photons contain based on their wavelength.

Wavelength of Light

The wavelength (\(\lambda\)) is the distance between successive peaks of a wave. It is usually measured in nanometers (nm) for light. In our scenario, the bulb emits light with a wavelength of 550 nm (or 550 x 10^-9 meters).
This particular wavelength falls within the visible spectrum that our eyes can detect, typically producing green light. Knowing the wavelength allows us to determine the photon's energy using other constants and formulas.

Joules per Second

A joule per second (J/s) is a unit of power. In this context, it measures how many joules of energy are emitted by the light bulb each second. Since we already know the power output of the bulb (75 W), and only 5% of it is converted to light, we see that 3.75 J/s is the light energy produced.
When only 10% of this light shines on a textbook, we calculate this as:
\[ 3.75 \text{ W} \times 0.10 = 0.375 \text{ J/s}. \]
This small amount of energy represents how much light energy the book is receiving every second.