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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

Expert verified
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

What are the de Broglie wavelengths of electrons with the following values of kinetic energy? (a) \(1.0 \mathrm{eV}\) (b) $1.0 \mathrm{keV} .
The distance between atoms in a crystal of \(\mathrm{NaCl}\) is $0.28 \mathrm{nm} .$ The crystal is being studied in a neutron diffraction experiment. At what speed must the neutrons be moving so that their de Broglie wavelength is \(0.28 \mathrm{nm} ?\)
A hydrogen atom has a radius of about \(0.05 \mathrm{nm}\) (a) Estimate the uncertainty in any component of the momentum of an electron confined to a region of this size. (b) From your answer to (a), estimate the electron's kinetic energy. (c) Does the estimate have the correct order of magnitude? (The ground-state kinetic energy predicted by the Bohr model is \(13.6 \mathrm{eV} .\) )
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