Problem 41
Question
Solar Power Photovoltaic cells convert solar energy into electricity. Could tantalum \(\left(\phi=6.81 \times 10^{-19} \mathrm{J}\right)\) be used to convert visible light to electricity? Assume that most of the electromagnetic energy from the Sun is in the visible region near \(500 \mathrm{nm}\)
Step-by-Step Solution
Verified Answer
Answer: Yes, tantalum can be used to convert visible light to electricity using solar power photovoltaic cells, since the energy of visible light ($$3.978 \times 10^{-19} \mathrm{J}$$) is greater than the work function of tantalum ($$6.81 \times 10^{-19} \mathrm{J}$$).
1Step 1: Find the energy of visible light
To find the energy of visible light, we can use the equation: $$E = \frac{hc}{\lambda}$$ where $$E$$ is the energy, $$h$$ is the Planck's constant ($$6.63 \times 10^{-34} \mathrm{Js}$$), $$c$$ is the speed of light ($$3\times 10^8 m/s$$), and $$\lambda$$ is the wavelength ($$500 \times 10^{-9} \mathrm{m}$$).
2Step 2: Calculate the energy of visible light
Plug the values into the equation: $$E = \frac{(6.63 \times 10^{-34} \mathrm{Js})(3 \times 10^8 \mathrm{m/s})}{500 \times 10^{-9} \mathrm{m}} $$
Now, calculate the energy: $$E = 3.978 \times 10^{-19} \mathrm{J}$$
3Step 3: Compare the energy with the work function of tantalum
Now that we have the energy of visible light, we can compare it to the work function of tantalum ($$\phi = 6.81 \times 10^{-19} \mathrm{J}$$). Since $$3.978 \times 10^{-19} \mathrm{J} > 6.81 \times 10^{-19} \mathrm{J}$$, the energy of the visible light is greater than the work function of tantalum.
4Step 4: Conclusion
Based on our calculations, tantalum can be used to convert visible light into electricity using solar power photovoltaic cells, since the energy of visible light is greater than the work function of tantalum.
Key Concepts
Solar EnergyVisible LightTantalum Work Function
Solar Energy
Solar energy is a renewable source of power derived from the sun's rays. This type of energy is crucial for sustainable development because it is both abundant and eco-friendly. Photovoltaic (PV) cells are devices that are designed to capture solar energy and convert it into electricity.
- Solar power is clean and does not emit harmful pollutants, making it a viable alternative to fossil fuels.
- It can be harnessed either directly using PV cells or indirectly by concentrating solar power systems.
- Solar energy technology has been advancing, leading to improved efficiency and lower costs.
Visible Light
Visible light is a small segment of the electromagnetic spectrum that is detectable by the human eye. It consists of wavelengths ranging approximately from 380 to 750 nanometers. When it comes to photovoltaic cells, visible light is of particular interest because it is abundant and contains sufficient energy to be converted into electricity.
- The energy of light can be calculated using the formula: \(E = \frac{hc}{\lambda}\), where \(E\) is the energy, \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength.
- The light near 500 nm, which is present in sunlight, is suitable for many photovoltaic applications.
- Understanding which parts of visible light can generate electricity helps in the design of efficient solar cells.
Tantalum Work Function
The work function of a material is the minimum amount of energy needed to remove an electron from the surface of that material. In the context of tantalum, the work function is essential in determining its suitability for use in photovoltaic cells.
- The work function of tantalum is \(6.81 \times 10^{-19} \mathrm{J}\).
- To convert light to electricity, the energy of the incoming photons must be greater than or equal to the work function.
- This comparison is critical in identifying materials for efficient energy conversion in PV cells.
Other exercises in this chapter
Problem 38
Which of the following have quantized values? Explain your selections. a. the pitch of a note played on a slide trombone b. the pitch of a note played on a flut
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Titanium \(\left(\phi=6.94 \times 10^{-19} \mathrm{J}\right)\) and silicon \((\phi=7.24 \times\) \(\left.10^{-19} \mathrm{J}\right)\) surfaces are irradiated wi
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With reference to Problem \(7.41,\) could tungsten \((\phi=7.20 \times\) \(\left.10^{-19} \mathrm{J}\right)\) be used to construct solar cells?
View solution Problem 43
The power of a red laser \((\lambda=630 \mathrm{nm})\) is 1.00 watt (abbreviated \(\mathrm{W}\), where \(1 \mathrm{W}=1 \mathrm{J} / \mathrm{s}\) ). How many ph
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