Problem 9 - College Physics

Question

The photoelectric threshold wavelength of a tungsten surface is \(272 \mathrm{nm} .\) (a) What are the threshold frequency and work function (in eV) of this tungsten? (b) Calculate the maximum kinetic energy (in eV) of the electrons ejected from this tungsten surface by ultraviolet radiation of frequency \(1.45 \times 10^{15} \mathrm{~Hz}\)

Step-by-Step Solution

Answer

Threshold frequency: \(1.10 \times 10^{15} \text{ Hz}\); Work function: 4.55 eV; Kinetic energy: 1.45 eV.

Step 1: Convert Wavelength to Frequency

First, let's find the threshold frequency using the formula: \( f = \frac{c}{\lambda} \), where \(c\) is the speed of light \( (3 \times 10^8 \text{ m/s})\) and \(\lambda\) is the wavelength. Convert \(272 \text{ nm} \) to meters: \(272 \text{ nm} = 272 \times 10^{-9} \text{ m}\). Hence, \( f = \frac{3 \times 10^8}{272 \times 10^{-9}} = 1.10 \times 10^{15} \text{ Hz} \).

Step 2: Calculate Work Function in eV

The work function \( \phi \) is calculated using the formula \( \phi = h \cdot f \), where \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{ J s}) \). First, find the energy in joules: \( \phi = 6.626 \times 10^{-34} \times 1.10 \times 10^{15} = 7.289 \times 10^{-19} \text{ J} \). Now, convert the energy from joules to electronvolts using the conversion 1 eV = \( 1.602 \times 10^{-19} \) J: \( \phi = \frac{7.289 \times 10^{-19}}{1.602 \times 10^{-19}} = 4.55 \text{ eV} \).

Step 3: Energy of Incident Photons

Calculate the energy of the incident photons with frequency \(1.45 \times 10^{15} \text{ Hz}\) using \( E = h \cdot f \). Thus, \( E = 6.626 \times 10^{-34} \cdot 1.45 \times 10^{15} = 9.604 \times 10^{-19} \text{ J} \). Convert this energy to eV: \( E = \frac{9.604 \times 10^{-19}}{1.602 \times 10^{-19}} = 6.00 \text{ eV} \).

Step 4: Calculate Maximum Kinetic Energy of Ejected Electrons

The maximum kinetic energy \( K_{max} \) of the electrons is the difference between the energy of the incident photons and the work function: \( K_{max} = E - \phi = 6.00 \text{ eV} - 4.55 \text{ eV} = 1.45 \text{ eV} \).

Key Concepts

Threshold Frequency

The Photoelectric Effect is a fantastic phenomenon that offers insights into quantum mechanics, where light can actually knock electrons out of a metal. In this context, the threshold frequency is crucial. This is the minimum frequency of incident light required to eject electrons from a material. Electrons will be emitted only when the incoming light has a frequency equal to or greater than this threshold frequency. For tungsten with a wavelength of 272 nm, we can calculate the threshold frequency using the formula: \( f = \frac{c}{\lambda} \). Given that \( c \) is the speed of light, approximately \( 3 \times 10^8 \text{ m/s} \), the conversion indicates the threshold frequency is indeed \( 1.10 \times 10^{15} \text{ Hz} \). This means that light, with a frequency less than this, will not have enough energy to cause electron emission.

Work Function

The work function is another pivotal concept in understanding the Photoelectric Effect. It refers to the minimum energy needed to remove an electron from the surface of a material. In essence, it's like a barrier that the light must overcome. For example, tungsten has a work function of 4.55 eV in this context. This value is derived by multiplying Planck's constant \( h \) by the threshold frequency. The calculation \( \phi = h \cdot f \) gives us a work function in joules, which we convert to electronvolts (eV). This value is significant because it determines how much energy incoming light must have to release electrons. Without sufficient energy to meet this work function, electrons will not be ejected.

Kinetic Energy of Electrons

Once the light's energy surpasses both the threshold frequency and the work function barrier, electrons are ejected with some kinetic energy. This energy is the excess left over after compensating for the work function. The formula to determine an electron's maximum kinetic energy is: \( K_{max} = E - \phi \), where \( E \) is the energy of the incoming photon and \( \phi \) is the work function. In this case, with ultraviolet radiation frequency \( 1.45 \times 10^{15} \text{ Hz} \), we calculate the photon energy first, and then subtract the work function. The remaining kinetic energy of 1.45 eV signifies how energetically the electrons will be ejected from the metal's surface.

Ultraviolet Radiation

Ultraviolet (UV) radiation is a type of electromagnetic radiation that has a higher frequency than visible light. In the Photoelectric Effect, UV radiation is often used due to its higher energy photons. This makes it effective at overcoming the work function of many materials. In our given scenario, the ultraviolet radiation with a frequency of \( 1.45 \times 10^{15} \text{ Hz} \) provides sufficient energy to not only eject electrons from tungsten but also leaves them with some kinetic energy. Understanding how UV radiation interacts with materials helps us not only in studying the Photoelectric Effect but also in various applications ranging from sterilization to the production of vitamin D. UV's high energy makes it especially useful in releasing electrons in these experimental setups.