Problem 11 - University Physics with Modern Physics

Question

When ultraviolet light with a wavelength of 400.0 \(\mathrm{nm}\) falls on a certain metal surface, the maximum kinetic energy of the emitted photoelectrons is measured to be 1.10 eV. What is the maximum kinetic energy of the photoelectrons when light of wavelength 300.0 nm falls on the same surface?

Step-by-Step Solution

Answer

The maximum kinetic energy of the photoelectrons for 300 nm light is greater than 1.10 eV.

Step 1: Convert Wavelength to Frequency

We start by converting the given wavelength of ultraviolet light into frequency. The speed of light (\(c\)) is \(3.00 \times 10^8\) m/s. Use the formula \(f = \frac{c}{\lambda}\), where \(\lambda\) is the wavelength. For 400.0 nm, \(\lambda = 400.0 \times 10^{-9}\) m. Thus, \(f_{400} = \frac{3.00 \times 10^8 \ m/s}{400.0 \times 10^{-9} \ m}\). Solve to find \(f_{400}\).

Step 2: Calculate Photon Energy for 400 nm Light

Using the frequency obtained from Step 1, calculate the energy of these photons using Planck's equation: \(E = h \cdot f\), where \(h\) is Planck’s constant \(6.626 \times 10^{-34}\) J·s. Thus, \(E_{400} = 6.626 \times 10^{-34} \cdot f_{400}\). Convert this energy from joules to electron volts (1 eV = \(1.602 \times 10^{-19}\) J).

Step 3: Determine the Work Function

The work function (\(\Phi\)) of the metal can be determined from the photon energy of the 400 nm light and its resulting kinetic energy: \(\Phi = E_{400} - KE_{400}\), where \(KE_{400} = 1.10\) eV.

Step 4: Calculate Frequency for 300 nm Light

Convert the wavelength for 300.0 nm to frequency using \(f = \frac{c}{\lambda}\). Here, \(\lambda = 300.0 \times 10^{-9}\) m. So, \(f_{300} = \frac{3.00 \times 10^8 \ m/s}{300.0 \times 10^{-9} \ m}\). Solve to get \(f_{300}\).

Step 5: Calculate Photon Energy for 300 nm Light

Using Planck's equation \(E = h \cdot f\), calculate the photon energy for the 300 nm light using \(f_{300}\) from Step 4: \(E_{300} = 6.626 \times 10^{-34} \cdot f_{300}\). Convert \(E_{300}\) from joules to electron volts.

Step 6: Calculate Maximum Kinetic Energy

The maximum kinetic energy for 300 nm light can be found using \(KE_{300} = E_{300} - \Phi\), where \(\Phi\) is from Step 3 and \(E_{300}\) is the energy for 300 nm light from Step 5.

Key Concepts

Photon Energy Calculation

Photon energy is the energy carried by a single photon, which is the basic quantum of light. To calculate this energy, we rely on Planck's equation: \( E = h \cdot f \), where:

• \( E \) is the photon energy.
• \( h \) is Planck’s constant \(6.626 \times 10^{-34}\) J·s.
• \( f \) is the frequency of the light.

The frequency \( f \) is linked to the wavelength \( \lambda \) of light using the formula \( f = \frac{c}{\lambda} \), with \( c \) representing the speed of light \(3.00 \times 10^8 \) m/s.
Substitute the wavelength into this equation to find the frequency.
Once the frequency is known, using Planck's equation lets us assess the photon's energy. Usually, this energy is expressed in joules, but it can also be converted to electron volts (eV), popular for atomic-scale measurements.
To switch from joules to electron volts, use the conversion \( 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \). This calculation helps understand how light of different wavelengths impacts materials, like in the photoelectric effect.

Ultraviolet Light

Ultraviolet (UV) light is a form of electromagnetic radiation with a wavelength shorter than visible light but longer than X-rays.
It ranges from about 10 nm to 400 nm and is known for its ability to affect certain materials differently, including causing chemical reactions and fluorescence.
For the photoelectric effect, UV light is particularly effective because its photons carry more energy due to their shorter wavelengths.
Shorter wavelengths mean higher frequencies, and therefore, higher photon energy.
This capability to provide high-energy photons makes UV light perfect for experiments where light needs to emit electrons from materials, such as metal surfaces.
Moreover, understanding the specific range and properties of ultraviolet light helps in determining its influence on photoelectric emissions.
In practical applications, ultraviolet light is explored not just in physics but also in fields like medicine and security, owing to its unique properties to affect organic and inorganic substances.

Maximum Kinetic Energy

In the context of the photoelectric effect, the maximum kinetic energy of emitted photoelectrons is an essential measure.
This energy describes the highest velocity with which electrons are ejected from a metal surface when irradiated by light.

• The kinetic energy depends on the energy of the incident photons minus the work function of the metal.
• The work function \( \Phi \) is the minimum energy needed to liberate an electron from the material.
• The equation connecting these quantities is \( KE_{\text{max}} = E_{\text{photon}} - \Phi \).

When a photon strikes a metal surface, its energy is divided into two parts: lifting the electron over the potential barrier (work function) and the remainder becoming the kinetic energy of the electron.
Hence, the maximum kinetic energy of photoelectrons differs based on both the incoming light's wavelength and the inherent properties of the metal.
Understanding this energy helps elaborate on the practical and theoretical aspects of photoelectron behavior, influencing technologies like photodetectors and solar cells.