/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 94 A technique called photo electro... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A technique called photo electron spectroscopy is used to measure the ionization energy of atoms. A sample is irradiated with ultraviolet (UV) light, and electrons are ejected from the valence shell. The kinetic energies of the ejected electrons are measured. Because the energy of the UV photon and the kinetic energy of the ejected electron are known, we can write \( h v=I E+\frac{1}{2} m u^{2} \) in which \(\nu\) is the frequency of the UV light, and \(m\) and \(u\) are the mass and velocity of the electron, respectively. In one experiment the kinetic energy of the ejected electron from potassium is found to be \(5.34 \times 10^{-19} \mathrm{~J}\) using a UV source of wavelength \(162 \mathrm{nm}\). Calculate the ionization energy of potassium. How can you be sure that this ionization energy corresponds to the electron in the valence shell (that is, the most loosely held electron)?

Short Answer

Expert verified
The ionization energy of potassium is 6.96 x 10^-19 J.

Step by step solution

01

Calculate the frequency of the UV light

The frequency (\(\nu\)) of the UV light can be calculated from the wavelength (\(\lambda\)) using the formula \(\nu = \frac{c}{\lambda}\), where \(c\) is the speed of light. Given that the speed of light \(c = 3 \times 10^8 \, m/s\) and the wavelength \(\lambda = 162 \times 10^{-9} \, m\) (162 nm), the frequency can be calculated as follows: \(\nu = \frac{3 \times 10^8 \, m/s}{162 \times 10^{-9} \, m} = 1.85 \times 10^{15}\, Hz\).
02

Calculate the energy of the incident UV photon

The energy of the incident UV photon can be calculated using the formula \(E = h\nu\), where \(h\) is Planck’s constant (\(6.63 \times 10^{-34} \, J \cdot s\)). The energy of the photon is: \(E = 6.63 \times 10^{-34}\, J \cdot s \times 1.85 \times 10^{15}\, Hz = 1.23 \times 10^{-18}\, J\).
03

Calculate the ionization energy

The ionization energy can be calculated by rearranging the given equation to: \(IE = h\nu - \frac{1}{2} mu^2\). Here, \(m\) and \(u\) are the mass and velocity of the electron, respectively, which can be replaced with the kinetic energy term: \(5.34 \times 10^{-19} \, J\). Now insert the calculated photon energy and given kinetic energy into this formula to get the ionization energy: \(IE = 1.23 \times 10^{-18}\, J - 5.34 \times 10^{-19}\, J = 6.96 \times 10^{-19}\, J\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ionization Energy
Ionization energy is a critical concept in photoelectron spectroscopy, and it refers to the energy needed to remove an electron from an atom or molecule. This energy is specific to each element and its electrons. In the context of photoelectron spectroscopy, we measure ionization energy by exposing atoms to high-energy light, like ultraviolet (UV) light, which provides enough energy to eject electrons from the atoms. When this happens, the electron absorbs energy from the UV light. If this energy exceeds the ionization energy, the electron is removed from the atom. The substantial concept here is that ionization energy relates to the strength with which an electron is bound to an atom. The more loosely bound the electron is, the less ionization energy is required to remove it. In the given exercise, by computing the kinetic energy of an ejected electron, and knowing the energy of the UV light used, one can deduce the ionization energy required to remove the electron from the atom. This calculation provides critical insights into the structure of the atom and how electrons are arranged around the nucleus.
Ultraviolet (UV) Light
Ultraviolet (UV) light is a form of electromagnetic radiation with a wavelength shorter than visible light but longer than X-rays. It holds a particular place in photoelectron spectroscopy due to the energy it carries. UV light photons possess enough energy to eject electrons from their atoms, which is why they are used in the process of measuring ionization energy. In the experiment provided, UV light with a specific wavelength interacts with potassium atoms. This wavelength of 162 nm (nanometers) translates into a frequency, which can be calculated using the speed of light. The frequency is then used to determine the energy of the photons in UV light. UV light's ability to knock electrons loose is what allows scientists to calculate the details of an atom's ionization energy. By knowing the energy that the UV photons contributed, and subsequently the kinetic energy gained by the ejected electron, the remaining energy—once subtracted by kinetic energy—is equal to the ionization energy. Thus, UV light acts as both the initiator and the measuring stick in this scientific process.
Kinetic Energy
Kinetic energy, in the realm of photoelectron spectroscopy, is the energy of motion that ejected electrons possess after they are dislodged from their parent atoms by the UV photons. It is a vital component in the equation used to calculate ionization energy, as it helps bridge the gap between photon energy and the energy required to remove the electron itself. The calculation of kinetic energy occurs when the ejected electrons' velocity is measured. Once you know the speed and mass of the electron, you can compute its kinetic energy using the relation: \[ \text{kinetic energy} = \frac{1}{2} m u^2 \] where \( m \) is the mass of the electron and \( u \) is its velocity.In the exercise, you know the specific kinetic energy (\(5.34 \times 10^{-19} \text{ J}\)) of the ejected electron from the experiment. This value is deducted from the total photon energy calculated using the UV light's frequency, and what's left represents the ionization energy needed to dislodge the electron in the first place. Thus, kinetic energy is a crucial element in piecing together the puzzle of ionization energy within an atom.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Indicate whether these elements exist as atomic species, molecular species, or extensive three dimensional structures in their most stable state at \(25^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm},\) and write the molecular or empirical formula for the elements: phosphorus, iodine, magnesium, neon, arsenic, sulfur, carbon, selenium, and oxygen.

Use the alkali metals and alkaline earth metals as examples to show how we can predict the chemical properties of elements simply from their electron configurations.

A student is given samples of three elements, \(X, Y\) and \(Z\), which could be an alkali metal, a member of Group \(4 \mathrm{~A},\) and a member of Group \(5 \mathrm{~A}\). She makes the following observations: Element X has a metallic luster and conducts electricity. It reacts slowly with hydrochloric acid to produce hydrogen gas. Element \(Y\) is a light-yellow solid that does not conduct electricity. Element \(Z\) has a metallic luster and conducts electricity. When exposed to air, it slowly forms a white powder. A solution of the white powder in water is basic. What can you conclude about the elements from these observations?

The formula for calculating the energies of an electron in a hydrogen-like ion is \( E_{n}=-\left(2.18 \times 10^{-18} \mathrm{~J}\right) Z^{2}\left(\frac{1}{n^{2}}\right) \) This equation cannot be applied to many-electron atoms. One way to modify it for the more complex atoms is to replace \(Z\) with \((Z-\sigma)\), in which \(Z\) is the atomic number and \(\sigma\) is a positive dimensionless quantity called the shielding constant. Consider the helium atom as an example. The physical significance of \(\sigma\) is that it represents the extent of shielding that the two 1 s electrons exert on each other. Thus, the quantity \((Z-\sigma)\) is appropriately called the "effective nuclear charge." Calculate the value of \(\sigma\) if the first ionization energy of helium is \(3.94 \times\) \(10^{-18} \mathrm{~J}\) per atom. (Ignore the minus sign in the given equation in your calculation.).

In each of the following pairs, indicate which one of the two species is smaller: (a) \(\mathrm{Cl}\) or \(\mathrm{Cl}^{-},\) (b) Na or \(\mathrm{Na}^{+}\), (c) \(\mathrm{O}^{2-}\) or \(\mathrm{S}^{2-},\) (d) \(\mathrm{Mg}^{2+}\) or \(\mathrm{Al}^{3+},\) (e) \(\mathrm{Au}^{+}\) or \(\mathrm{Au}^{3+}\).

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.