Question

In order for a photon of light incident upon metallic potassium to eliminate an electron from it, the photon must have a minimum energy of \(4.33 \mathrm{eV}\) (photoelectric work function for potassium). What is the wavelength of a photon of this energy? Assume \(\mathrm{h}=6.626 \times 10^{-27}\) erg-sec and \(c=2.998 \times 10^{10} \mathrm{~cm} / \mathrm{sec}\)

Short answer

The wavelength of a photon with enough energy to eliminate an electron from metallic potassium is approximately 286.5 nm.

Step-by-step solution

Write down the given information

The photoelectric work function of potassium: \(E = 4.33 eV\) Planck's constant (\(h\)): \(6.626\times10^{-27} erg\cdot s\) Speed of light (\(c\)): \(2.998\times10^{10} \frac{cm}{s}\)

Convert energy from eV to ergs

First, we need to convert the energy from electron volts (eV) to ergs, since Planck's constant is given in ergs. 1 electron volt (\(eV\)) \(\approx 1.602\times10^{-12} erg\) So, \(E = 4.33 eV \times \frac{1.602\times10^{-12} erg}{1 eV} \approx 6.936 \times 10^{-12} erg\)

Use the formula for energy in terms of wavelength

Now we can use the formula for energy in terms of wavelength and Planck's constant: \(E = \frac{hc}{\lambda}\) Where, \(E\) = Energy (in ergs) \(h\) = Planck's constant (in ergs·s) \(c\) = Speed of light (in cm/s) \(\lambda\) = Wavelength (in cm)

Solve for wavelength

We can now rearrange the formula to find the wavelength (\(\lambda\)): \(\lambda = \frac{hc}{E}\) Plugging the values, \(\lambda = \frac{(6.626\times10^{-27} erg\cdot s)(2.998\times10^{10}\frac{cm}{s})}{6.936\times10^{-12}erg}\) Solve for \(\lambda\): \(\lambda \approx 2.865\times10^{-5}cm\)

Convert the wavelength to nanometers

Finally, we can convert the wavelength from centimeters to nanometers for a more practical result. 1 cm = \(10^7\) nm \(\lambda \approx 2.865\times10^{-5}cm \times \frac{10^7 nm}{1 cm} \approx 286.5 nm\)

Conclusion

The wavelength of a photon with enough energy to eliminate an electron from metallic potassium is approximately 286.5 nm.

Key concepts

Planck's Constant

Planck's constant, denoted as \( h \), is a fundamental constant in quantum mechanics. It has a value of \(6.626 \times 10^{-27}\) erg-seconds. This constant is crucial in calculating the energy of photons, which are particles of light. Planck's constant provides the proportionality factor that relates the energy of a photon to its frequency. The formula used is \( E = h u \), where \( E \) is the energy and \( u \) is the frequency of the photon.

• Planck's constant is a small value, showcasing the tiny scale at which quantum mechanics operates.
• This constant is essential in understanding phenomena such as the photoelectric effect.

Planck’s constant helps explain how energy is quantized in discrete amounts, changing forever how we view light and energy.

Wavelength Calculation

The wavelength of a photon is calculated using the formula \( E = \frac{hc}{\lambda} \), where \( E \) is energy, \( h \) is Planck’s constant, \( c \) is the speed of light, and \( \lambda \) is the wavelength. To find the wavelength, we rearrange the formula to \( \lambda = \frac{hc}{E} \).
This calculation allows us to determine how energy and wavelength are interrelated.

• This formula demonstrates the inverse relationship between energy and wavelength: as the energy increases, the wavelength decreases.
• Understanding this relationship is crucial in physics and chemistry, especially when examining light and electromagnetic radiation.

In the provided exercise, calculating the wavelength of a photon starts with converting energy from electron volts to ergs, ensuring unit consistency for accurate results.

Energy Conversion

In physics, energy conversion is often necessary when different units are used. In the context of the photoelectric effect, energy is sometimes given in electron volts (eV), but calculations might require conversions to other units such as ergs.

• 1 electron volt is equivalent to \(1.602 \times 10^{-12}\) ergs, a conversion crucial for precise calculations.
• Energy conversion is necessary for using Planck’s constant when it's provided in ergs-second units.

This process ensures the energy used in equations is consistent with the units of other constants, facilitating accurate and reliable outcomes in physics problems.

Photon Energy

Photon energy is an essential concept in understanding light as a particle. It signifies the energy carried by a single photon and can be calculated using the formula \( E = h u \) or \( E = \frac{hc}{\lambda} \), depending on whether frequency or wavelength is utilized.

• Photon energy is measured in electron volts (eV) or ergs, reflecting the photon's power to cause an effect, such as ejecting an electron in the photoelectric effect.
• The amount of photon energy determines if it can overcome the work function of a material to emit electrons.

Understanding photon energy helps explain how light interacts with matter, laying the groundwork for technologies like solar cells and photo detectors that rely on photon interactions.