Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

To resolve details of an object, you must use a wavelength that is about the same size, or smaller, than the details you want to observe. Suppose you want to study a molecule that is about \(1.000 \times 10^{-10} \mathrm{m}\) in length. (a) What minimum photon energy is required to study this molecule? (b) What is the minimum kinetic energy of electrons that could study this? (c) Through what potential difference should the electrons be accelerated to reach this energy?

Short Answer

Expert verified
Answer: (a) The minimum photon energy required to study the molecule is \(1.987 \times 10^{-25}\mathrm{J}\). (b) The minimum kinetic energy of electrons required to study the molecule is \(2.426 \times 10^{-18}\mathrm{J}\). (c) The potential difference required to accelerate the electrons to the minimum kinetic energy is \(15.16\mathrm{V}\).

Step by step solution

01

Find the minimum wavelength for studying the molecule

To study an object of length \(1.000 \times 10^{-10} \mathrm{m}\), we need a wavelength smaller or equal to the length of the object. So, the minimum wavelength (\(\lambda_{min}\)) required is: \(\lambda_{min} = 1.000 \times 10^{-10}\mathrm{m}\)
02

Calculate the minimum photon energy

To find the minimum photon energy, we can use the formula relating energy (\(E\)) and wavelength (\(\lambda\)): \(E = \dfrac{h c}{\lambda}\) where \(h\) is the Planck's constant (\(6.626 \times 10^{-34}\mathrm{Js}\)), \(c\) is the speed of light (\(3.0 \times 10^8\mathrm{m/s}\)), and \(\lambda\) is the minimum wavelength calculated in step 1. Substitute the given values and calculate the minimum photon energy: \(E = \dfrac{(6.626 \times 10^{-34}\mathrm{Js})(3.0 \times 10^8\mathrm{m/s})}{1.000 \times 10^{-10}\mathrm{m}} = 1.987 \times 10^{-25}\mathrm{J}\) So, the minimum photon energy required to study the molecule is \(1.987 \times 10^{-25}\mathrm{J}\).
03

Calculate the minimum electron kinetic energy

Since we are using electrons to study the molecule, we can use the de Broglie wavelength formula to find the minimum kinetic energy required. The formula for de Broglie wavelength is: \(\lambda = \dfrac{h}{p}\) Here, \(p\) represents the momentum of the electron. To relate momentum to kinetic energy, we can use the formula: \(p = \sqrt{2 m_e K}\) where \(m_e\) is the mass of an electron (\(9.11 \times 10^{-31}\mathrm{kg}\)) and \(K\) is the kinetic energy. Combine these two equations to find the minimum kinetic energy: \(\lambda_{min} = \dfrac{h}{\sqrt{2 m_e K}}\) Solving for \(K\): \(K = \dfrac{h^2}{2m_e\lambda_{min}^2}\) Substitute the given values and calculate the minimum kinetic energy: \(K = \dfrac{(6.626 \times 10^{-34}\mathrm{Js})^2}{2(9.11 \times 10^{-31}\mathrm{kg})(1.000 \times 10^{-10}\mathrm{m})^2} = 2.426 \times 10^{-18}\mathrm{J}\) So, the minimum kinetic energy of electrons required to study the molecule is \(2.426 \times 10^{-18}\mathrm{J}\).
04

Calculate the potential difference for accelerating the electrons

We can calculate the potential difference (\(V\)) needed to achieve the required kinetic energy using the formula: \(K = eV\) where \(e\) is the elementary charge (\(1.6 \times 10^{-19}\mathrm{C}\)). Solve for \(V\): \(V = \dfrac{K}{e}\) Substitute the calculated minimum kinetic energy and find the potential difference: \(V = \dfrac{2.426 \times 10^{-18}\mathrm{J}}{1.6 \times 10^{-19}\mathrm{C}} = 15.16\mathrm{V}\) So, the potential difference required to accelerate the electrons to the minimum kinetic energy is \(15.16\mathrm{V}\). In summary: (a) The minimum photon energy required to study the molecule is \(1.987 \times 10^{-25}\mathrm{J}\). (b) The minimum kinetic energy of electrons required to study the molecule is \(2.426 \times 10^{-18}\mathrm{J}\). (c) The potential difference required to accelerate the electrons to the minimum kinetic energy is \(15.16\mathrm{V}\).

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 Vaia!

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

What is the ratio of the wavelength of a 0.100-keV photon to the wavelength of a 0.100-keV electron?
A scanning electron microscope is used to look at cell structure with 10 -nm resolution. A beam of electrons from a hot filament is accelerated with a voltage of \(12 \mathrm{kV}\) and then focused to a small spot on the specimen. (a) What is the wavelength in nanometers of the beam of incoming electrons? (b) If the size of the focal spot were determined only by diffraction, and if the diameter of the electron lens is one fifth the distance from the lens to the specimen, what would be the minimum separation resolvable on the specimen? (In practice, the resolution is limited much more by aberrations in the magnetic lenses and other factors.)
An image of a biological sample is to have a resolution of \(5 \mathrm{nm} .\) (a) What is the kinetic energy of a beam of electrons with a de Broglie wavelength of \(5.0 \mathrm{nm} ?\) (b) Through what potential difference should the electrons be accelerated to have this wavelength? (c) Why not just use a light microscope with a wavelength of 5 nm to image the sample?
(a) Show that the number of electron states in a subshell is \(4 \ell+2 .\) (b) By summing the number of states in each of the subshells, show that the number of states in a shell is \(2 n^{2} .\) [Hint: The sum of the first \(n\) odd integers, from 1 to \(2 n-1,\) is \(n^{2} .\) That comes from regrouping the sum in pairs, starting by adding the largest to the smallest: \(1+3+5+\dots+(2 n-5)+(2 n-3)+(2 n-1)\) \(=[1+(2 n-1)]+[3+(2 n-3)]+[5+(2 n-5)]+\cdots\) \(=2 n+2 n+2 n+\cdots=2 n \times \frac{n}{2}=n^{2}\)
The energy-time uncertainty principle allows for the creation of virtual particles, that appear from a vacuum for a very brief period of time $\Delta t,\( then disappear again. This can happen as long as \)\Delta E \Delta t=\hbar / 2,\( where \)\Delta E$ is the rest energy of the particle. (a) How long could an electron created from the vacuum exist according to the uncertainty principle? (b) How long could a shot put with a mass of \(7 \mathrm{kg}\) created from the vacuum exist according to the uncertainty principle?
See all solutions

Recommended explanations on Physics 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.

Sign-up for free