Question

Only a fraction of the electric energy supplied to a tungsten lightbulb is converted to visible light. The rest of the energy shows up as infrared radiation (i.e., 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

Approximately \(3.12 \times 10^{19}\) photons are emitted per second.

Step-by-step solution

Calculate the energy used for visible light

The 75-W lightbulb converts 15% of its energy to visible light. We find the power of visible light by multiplying the total power by the conversion efficiency: \[ P_{visible} = 0.15 \times 75 \text{ W} = 11.25 \text{ W} \] This means 11.25 J/s is used for visible light.

Calculate the energy of a single photon

Each photon has energy given by \[ E = \frac{hc}{\lambda} \] where \( h = 6.626 \times 10^{-34} \text{ J}\cdot\text{s} \) (Planck's constant), \( c = 3 \times 10^8 \text{ m/s} \) (speed of light), and \( \lambda = 550 \times 10^{-9} \text{ m} \) (wavelength).Substituting the values:\[ E = \frac{6.626 \times 10^{-34} \cdot 3 \times 10^8}{550 \times 10^{-9}} \approx 3.61 \times 10^{-19} \text{ J} \]

Calculate the number of photons emitted per second

To find the number of photons emitted per second, divide the power used for visible light by the energy of one photon:\[ n = \frac{P_{visible}}{E} = \frac{11.25 \text{ J/s}}{3.61 \times 10^{-19} \text{ J/photon}} \approx 3.12 \times 10^{19} \text{ photons/second} \] Therefore, approximately \(3.12 \times 10^{19}\) photons are emitted every second.

Key concepts

Visible Light Conversion

When we talk about **visible light conversion** in light bulbs, we refer to the process where electrical energy is transformed into light energy that we can see. However, not all the electrical energy in a light bulb is used efficiently for this purpose. For example, a tungsten lightbulb might emit a significant portion of its energy as heat or infrared light, which we can't see.

• For a 75-Watt tungsten light bulb, only a small percentage, say 15%, is converted into visible light.
• This means that out of every 75 watts provided, only 11.25 watts turn into visible light.

Understanding this inefficiency helps us appreciate why some bulbs, like LEDs, are more preferred due to better energy efficiency, converting more electricity into visible light instead of heat.

Planck's Constant

The role of **Planck's constant** in this context is critical because it connects the concept of light as both a wave and a particle.
Planck's constant (

h = 6.626 \times 10^{-34} \text{ J}\cdot\text{s}

) is used to calculate the energy of photons, which are tiny packets of light energy.
Every photon can't possess just any amount of energy; its energy is specifically determined by the equation:\[ E = \frac{hc}{\lambda} \] This shows that the energy of a single photon is proportionally related to Planck's constant and inversely related to the light's wavelength.

Wavelength Calculation

**Wavelength calculation** involves measuring the distance between two consecutive peaks of a wave. For visible light, wavelengths are typically measured in nanometers (nm).
In visible light, the wavelength determines the color. In our case, the light's wavelength is 550 nm, which sits within the green/yellow spectrum of visible light.

• The relationship between the wavelength and energy of a photon is inverse. A shorter wavelength (like blue) means more energy, while a longer wavelength (like red) means less energy.
• This is captured in the formula, \( E = \frac{hc}{\lambda} \), where the energy

  E

  of a photon is calculated using the wavelength (\( \lambda \)).

Understanding wavelength calculations helps in designing lightbulbs that maximize desirable wavelengths for specific lighting needs.

Energy Efficiency of Lightbulbs

The **energy efficiency of lightbulbs** indicates how well a lightbulb converts electrical energy into visible light.
Traditional tungsten lightbulbs are less efficient because they waste a lot of energy as heat rather than light.

• The efficiency of a lightbulb can be calculated by looking at the ratio of visible light output to the total energy input.
• In our exercise, a 75-W bulb emits only 15% of its energy as visible light, which is 11.25 watts.

LEDs and other modern lighting options have much higher efficiency, reducing energy waste significantly, which is why they are more environmentally friendly and cost-effective in the long run.