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Chapter 28: Problem 57

A magnesium ion \(\mathrm{Mg}^{2+}\) is accelerated through a potential difference of \(22 \mathrm{kV}\). What is the de Broglie wavelength of this ion?

Short Answer

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Question: Calculate the de Broglie wavelength of a magnesium ion (\(\mathrm{Mg}^{2+}\)) accelerated through a potential difference of \(22\,\text{kV}\). Answer: To find the de Broglie wavelength of the magnesium ion, calculate the kinetic energy, speed, and finally the wavelength using the following equations: 1. \(KE = 2e(22\times10^3\,\text{V})\) 2. \(v = \sqrt{\frac{2KE}{m}}\) 3. \(\lambda = \frac{h}{mv}\) Substitute the known values for the constants, mass, and speed of the magnesium ion to obtain the de Broglie wavelength: \(\lambda = \frac{h}{\sqrt{\frac{2(2e)(22\times10^3\,\text{V})}{\frac{M}{N_A}}}}\)

Step by step solution

01

Calculate the kinetic energy of the magnesium ion

To calculate the kinetic energy of the ion after it has been accelerated through a potential difference of \(22\,\text{kV}\), we'll use the formula \(KE = qV\). Here, \(q\) is the charge of the ion, and \(V\) is the potential difference. The charge of a magnesium ion is +2 times the elementary charge, so \(q = 2e\). The potential difference is given in kilovolts, so we'll convert it to volts: \(V =22\times10^3\,\text{V}\). Thus, the kinetic energy is: \(KE = 2e(22\times10^3\,\text{V})\)
02

Find the speed of the magnesium ion

To find the speed of the ion after it has gained kinetic energy, we'll use the formula \(KE = \frac{1}{2}mv^2\), where \(m\) is the mass of the ion and \(v\) is its speed. We can rearrange the formula to solve for \(v\): \(v = \sqrt{\frac{2KE}{m}}\) The mass of a magnesium ion can be obtained using its molar mass and Avogadro's number: \(m = \frac{M}{N_A}\), where \(M\) is the molar mass (approximately \(24.3\times10^{-3}\,\mathrm{kg/mol}\)) and \(N_A\) is Avogadro's number (\(6.022\times10^{23}\,\mathrm{mol^{-1}}\)). Thus, the speed of the ion is: \(v = \sqrt{\frac{2(2e)(22\times10^3\,\text{V})}{\frac{M}{N_A}}}\)
03

Calculate the de Broglie wavelength of the magnesium ion

Now that we have the speed of the ion, we can find its de Broglie wavelength using the formula \(\lambda = \frac{h}{mv}\), where \(\lambda\) is the wavelength, \(h\) is Planck's constant (\(6.626\times10^{-34}\,\text{J}\cdot\text{s}\)), and \(m\) and \(v\) are the mass and speed of the ion, found in the previous step. Plugging in the values, the de Broglie wavelength is: \(\lambda = \frac{h}{\sqrt{\frac{2(2e)(22\times10^3\,\text{V})}{\frac{M}{N_A}}}}\)

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Most popular questions from this chapter

An electron is confined to a one-dimensional box of length \(L\). When the electron makes a transition from its first excited state to the ground state, it emits a photon of energy 0.20 eV. (a) What is the ground-state energy (in electron-volts) of the electron in this box? (b) What are the energies (in electron-volts) of the photons that can be emitted when the electron starts in its third excited state and makes transitions downwards to the ground state (either directly or through intervening states)? (c) Sketch the wave function of the electron in the third excited state. (d) If the box were somehow made longer, how would the electron's new energy level spacings compare with its old ones? (Would they be greater, smaller, or the same? Or is more information needed to answer this question? Explain.)
The particle in a box model is often used to make rough estimates of energy level spacings. Suppose that you have a proton confined to a one-dimensional box of length equal to a nuclear diameter (about \(10^{-14} \mathrm{m}\) ). (a) What is the energy difference between the first excited state and the ground state of this proton in the box? (b) If this energy is emitted as a photon as the excited proton falls back to the ground state, what is the wavelength and frequency of the electromagnetic wave emitted? In what part of the spectrum does it lie? (c) Sketch the wave function \(\psi\) as a function of position for the proton in this box for the ground state and each of the first three excited states.
Before the discovery of the neutron, one theory of the nucleus proposed that the nucleus contains protons and electrons. For example, the helium-4 nucleus would contain 4 protons and 2 electrons instead of - as we now know to be true- 2 protons and 2 neutrons. (a) Assuming that the electron moves at nonrelativistic speeds, find the ground-state energy in mega-electron- volts of an electron confined to a one-dimensional box of length $5.0 \mathrm{fm}\( (the approximate diameter of the \)^{4} \mathrm{He}$ nucleus). (The electron actually does move at relativistic speeds. See Problem \(80 .)\) (b) What can you conclude about the electron-proton model of the nucleus? The binding energy of the \(^{4} \mathrm{He}\) nucleus - the energy that would have to be supplied to break the nucleus into its constituent particles-is about \(28 \mathrm{MeV} .\) (c) Repeat (a) for a neutron confined to the nucleus (instead of an electron). Compare your result with (a) and comment on the viability of the proton-neutron theory relative to the electron-proton theory.
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 double-slit interference experiment is performed with 2.0-ev photons. The same pair of slits is then used for an experiment with electrons. What is the kinetic energy of the electrons if the interference pattern is the same as for the photons (i.e., the spacing between maxima is the same)?
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