Problem 50 - Chemistry An Atoms First Approach

Question

It takes \(208.4 \mathrm{kJ}\) of energy to remove 1 mole of electrons from an atom on the surface of rubidium metal. (1 mol electrons = \(6.022 \times 10^{23}\) electrons. How much energy does it take to remove a single electron from an atom on the surface of solid rubidium? What is the maximum wavelength of light capable of doing this?

Step-by-Step Solution

Answer

The energy required to remove a single electron from a rubidium atom is \(3.46\times10^{-19}\,\mathrm{J/electron}\). The maximum wavelength of light capable of providing this energy is \(5.73\times10^{-7}\,\mathrm{m}\) or \(573\,\mathrm{nm}\).

Step 1: (Step 1: Calculate energy required to remove a single electron from a rubidium atom)

We are given that it takes \(208.4\,\mathrm{kJ/mol}\) of energy to remove one mole of electrons from rubidium atoms. Given that 1 mole of electrons comprises \(6.022 \times 10^{23}\) electrons, we can calculate the energy required to remove a single electron as follows: Energy per electron \(=\frac{Total\,energy\,to\,remove\,one\,mole\,of\,electrons}{Number\,of\,electrons\,in\,one\,mole}\) \(=\frac{208.4\,\mathrm{kJ/mol}}{6.022\times10^{23}\,\mathrm{electrons/mol}}\) It is more convenient to work with energy in units of Joules (J), so first, convert kJ to J: \(208.4\,\mathrm{kJ/mol} \times \frac{1000\,\mathrm{J}}{1\,\mathrm{kJ}}=208400\,\mathrm{J/mol}\) Energy per electron \(=\frac{208400\,\mathrm{J/mol}}{6.022\times10^{23}\,\mathrm{electrons/mol}}=3.46\times10^{-19}\,\mathrm{J/electron}\)

Step 2: (Step 2: Determine the maximum wavelength of light capable of providing this energy)

To find the maximum wavelength capable of providing the required energy to remove a single electron, we will use the energy-wavelength relationship for light, which is given by the equation: \(E = \frac{hc}{\lambda}\) Where: E is the energy (in Joules), h is the Planck's constant \(=6.63 \times 10^{-34}\,\mathrm{J\cdot s}\), c is the speed of light \(=3.00\times10^{8}\,\mathrm{m/s}\), and \(\lambda\) is the wavelength (in meters). Rearrange the equation to solve for the wavelength: \(\lambda = \frac{hc}{E}\) Plug in the values for the constants and the energy determined in step 1: \(\lambda = \frac{(6.63 \times 10^{-34}\,\mathrm{J\cdot s})(3.00\times10^{8}\,\mathrm{m/s})}{3.46\times10^{-19}\,\mathrm{J}}=5.73\times10^{-7}\,\mathrm{m}\) So, the maximum wavelength of light capable of removing a single electron from a rubidium atom is \(5.73\times10^{-7}\,\mathrm{m}\) or \(573\,\mathrm{nm}\).

Key Concepts

Energy Conversion

The concept of energy conversion plays a crucial role in the study of photoelectric effects and various chemical processes. In the context of removing electrons from an atom, energy conversion involves transferring a specific amount of energy to the electrons to overcome the binding forces holding them in place. This is typically measured in kilojoules per mole (kJ/mol), but can be more useful in smaller measurements like Joules (J) when dealing with single electrons since the individual energy amount is quite small. For instance, in the removal of electrons from rubidium, we start with a bulk measurement, knowing that an energy of 208.4 kJ/mol is required, which applies to one complete mole of electrons. To find the energy for a single electron, the total energy of 208.4 kJ is converted to J, giving 208,400 J, and then divided by Avogadro's number, approximately \(6.022 \times 10^{23}\). This calculation results in an energy conversion showing each electron's required energy as \(3.46 \times 10^{-19}\) J, making it easier to understand and apply in subsequent calculations like determining the wavelength of light capable of triggering electron ejection. Such energy conversion not only facilitates more manageable calculations on a small scale but also provides a clearer understanding of the atomic and sub-atomic forces at play.

Planck's Constant

Planck's constant is a fundamental value in quantum mechanics that bridges the gap between the macroscopic and quantum worlds. It is integral to understanding energy relationships in the photoelectric effect. Often denoted as \( h \), Planck's constant is commonly used in formulas that involve the energy of photons, particularly in relation to frequency or wavelength when discussing electromagnetic radiation.The accepted value for Planck's constant is \(6.63 \times 10^{-34}\, \mathrm{J\cdot s}\), representing a very tiny but universally significant number that appears throughout quantum physics. This constant is used to convert the frequency of electromagnetic radiation into energy. In practical applications, such as calculating the maximum wavelength of light capable of inducing electron ejection in rubidium, Planck's constant enables us to relate a specific energy amount to a wavelength using the equation \( E = \frac{hc}{\lambda} \). By rearranging this equation to solve for wavelength, or \( \lambda \), it showcases how Planck's constant is pivotal for linking energy (from step 1) with light properties, allowing us to derive the wavelength, giving insight into the type of light necessary to produce such an energy exchange.

Wavelength Calculation

Calculating wavelength in the context of the photoelectric effect connects the energy of a photon to its ability to dislodge electrons from a material such as rubidium. This calculation is crucial because it ties together various physical constants and properties to better understand the overall process.Using the energy value per electron calculated from converting the bulk mole energy, \(3.46 \times 10^{-19}\) J, we can determine the maximum wavelength of light capable of removing an electron. The relation used is \( E = \frac{hc}{\lambda} \), which elegantly combines Planck's constant (\( h = 6.63 \times 10^{-34}\, \mathrm{J\cdot s}\)), the speed of light (\( c = 3.00 \times 10^{8}\, \mathrm{m/s}\)), and the identified energy.To find the wavelength \( \lambda \), rearrange the formula: \( \lambda = \frac{hc}{E} \). Plugging in the known values gives \( \lambda = \frac{(6.63 \times 10^{-34}\, \mathrm{J\cdot s})(3.00 \times 10^{8}\, \mathrm{m/s})}{3.46 \times 10^{-19}\, \mathrm{J}}\), resulting in a wavelength of \(573\, \mathrm{nm}\).This wavelength calculation shows what is practically needed for light to effectively remove an electron, providing insight into the nature of such electromagnetic radiation involved in the photoelectric effect.