Problem 44 - Chemistry: An Atoms-Focused Approach

Question

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 with UV radiation with a wavelength of \(250 \mathrm{nm}\). Which surface emits clectrons with the longer wavelength? What is the wavelength of the electrons emitted by the titanium surface?

Step-by-Step Solution

Answer

And what is the wavelength of electrons emitted by the titanium surface? Answer: The silicon surface emits electrons with a longer wavelength. The wavelength of electrons emitted by the titanium surface is approximately \(2.75 \times 10^{-10} \mathrm{m}\).

Step 1: Find the energy of the incident UV photons

We are given the wavelength of the UV radiation, which is 250 nm. First, we need to convert it to meters: \(\lambda_\text{UV} = 250 \times 10^{-9} \mathrm{m}\) The energy of a photon can be calculated using the formula: \(E_\text{photon} = \dfrac{hc}{\lambda_\text{UV}}\) where: \(E_\text{photon}\): Energy of the photon \(h\): Planck's constant \((h = 6.63 \times 10^{-34} \mathrm{Js})\) \(c\): Speed of light \((c = 3 \times 10^{8} \mathrm{m\,s^{-1}})\) \(\lambda_\text{UV}\): Wavelength of the UV radiation Substitute the values into the formula and calculate the energy of the incident UV photons. \(E_\text{photon} = \dfrac{6.63 \times 10^{-34} \mathrm{Js} \times 3 \times 10^{8} \mathrm{m\,s^{-1}}}{250 \times 10^{-9} \mathrm{m}} = 7.98 \times 10^{-19} \mathrm{J}\)

Step 2: Calculate the maximum kinetic energy of emitted electrons for each material

According to the photoelectric effect, the maximum kinetic energy of the emitted electron, \(K_\text{max}\), can be found by: \(K_\text{max} = E_\text{photon} - \Phi\) where: \(\Phi\): Work function of the material For titanium, \(K_\text{max,\,Ti} = 7.98 \times 10^{-19} \mathrm{J} - 6.94 \times 10^{-19} \mathrm{J} = 1.04 \times 10^{-19} \mathrm{J}\) For silicon, \(K_\text{max,\,Si} = 7.98 \times 10^{-19} \mathrm{J} - 7.24 \times 10^{-19} \mathrm{J} = 0.74 \times 10^{-19} \mathrm{J}\)

Step 3: Find the de Broglie wavelength of emitted electrons

To find the de Broglie wavelength of emitted electrons, we use the formula: \(\lambda_\text{e} = \dfrac{h}{\sqrt{2m_\text{e} K_\text{max}}}\) where: \(\lambda_\text{e}\): De Broglie wavelength of the emitted electron \(m_\text{e}\): Electron mass \((m_\text{e} = 9.11 \times 10^{-31} \mathrm{kg})\) For titanium, \(\lambda_\text{e,Ti} = \dfrac{6.63 \times 10^{-34} \mathrm{Js}}{\sqrt{2 \times 9.11 \times 10^{-31} \mathrm{kg} \times 1.04 \times 10^{-19} \mathrm{J}}} \approx 2.75 \times 10^{-10} \mathrm{m}\) For silicon, \(\lambda_\text{e,Si} = \dfrac{6.63 \times 10^{-34} \mathrm{Js}}{\sqrt{2 \times 9.11 \times 10^{-31} \mathrm{kg} \times 0.74 \times 10^{-19} \mathrm{J}}} \approx 3.20 \times 10^{-10} \mathrm{m}\)

Step 4: Compare the wavelengths of emitted electrons and find the answer

Now, we have the de Broglie wavelengths of the emitted electrons for both titanium and silicon: \(\lambda_\text{e,Ti} \approx 2.75 \times 10^{-10} \mathrm{m}\) \(\lambda_\text{e,Si} \approx 3.20 \times 10^{-10} \mathrm{m}\) Since \(\lambda_\text{e,Si} > \lambda_\text{e,Ti}\), we conclude that the silicon surface emits electrons with a longer wavelength. The wavelength of the electrons emitted by the titanium surface is approximately \(2.75 \times 10^{-10} \mathrm{m}\).

Key Concepts

De Broglie Wavelength

The de Broglie wavelength is a fundamental concept in quantum mechanics that describes the wave nature of particles. According to this theory, every particle has an associated wavelength that depends on its momentum. The equation for the de Broglie wavelength is:
\[\lambda = \frac{h}{p}\]where \(\lambda\) is the de Broglie wavelength, \(h\) is Planck's constant, and \(p\) is the momentum of the particle.

The concept is crucial when analyzing particles like electrons that exhibit both wave-like and particle-like properties. In the context of the photoelectric effect, electrons emitted from a surface under light exposure show de Broglie wavelengths that can be calculated using the kinetic energy they possess post-emission.

• It helps us understand the nature of particles smaller than atoms.
• The wavelength is inversely proportional to the kinetic energy of the electron.
• Important for predicting how particles like electrons will behave in quantum mechanics.

For example, for an electron emitted by titanium after absorbing UV light, the de Broglie wavelength is given by:
\[\lambda_\text{e} = \frac{h}{\sqrt{2m_\text{e} K_\text{max}}}\]This formula highlights the linkage between kinetic energy and the wavelength of an electron. It allows scientists to study the particle nature of electrons more deeply.

Work Function

The work function is the minimum energy required to remove an electron from a solid's surface. It is a critical concept in the study of the photoelectric effect, as it determines how easily electrons can be emitted from a material when light is shone on it.

Each material has a unique work function represented by \(\Phi\). It is measured in joules and is intrinsic to the material's electronic structure.

• Lower work function implies electrons can be emitted at lower energy.
• Varies across different materials.
• Higher work function materials require higher energy photons to emit electrons.

In the example problem, titanium has a work function of \(6.94 \times 10^{-19} \mathrm{J}\), while silicon's work function is \(7.24 \times 10^{-19} \mathrm{J}\). The photon energy from UV light must be greater than these work functions to emit electrons, as illustrated by the photoelectric equation:
\[K_\text{max} = E_\text{photon} - \Phi\]where \(K_\text{max}\) is the maximum kinetic energy of the emitted electrons. By comparing the work functions, one can predict which material will emit electrons with more ease.

Kinetic Energy of Electrons

The kinetic energy of electrons emitted in the photoelectric effect is a direct outcome of the energy balance after absorbing photons. It tells us how much energy the ejected electrons have after overcoming the work function.

The maximum kinetic energy \(K_\text{max}\) is given by:
\[K_\text{max} = E_\text{photon} - \Phi\]where \(E_\text{photon}\) is the energy of the incoming photon, and \(\Phi\) is the work function of the material.

• The equation shows the remaining energy after the work function has been overcome.
• Higher photon energy means the potential for higher kinetic energy in electrons.
• Varies based on material and incident light energy.

For the titanium surface in the problem, \(K_\text{max}\) was calculated to be \(1.04 \times 10^{-19} \mathrm{J}\). This indicates that titanium's electrons have this much kinetic energy available post-emission when irradiated with the given UV light. Understanding this helps in applications like photoelectron spectroscopy, where kinetic energy distributions are critical for analyzing material properties.