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Chapter 17: Problem 2348

An electron is at a distance of \(10 \mathrm{~m}\) form a charge of $10 \mathrm{C}\(. Its total energy is \)15.6 \times 10^{-10} \mathrm{~J}$. Its de Broglie wavelength at this point is \(\ldots \ldots\) $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg} . \mathrm{K}=9 \times 10^{9} \mathrm{SI}\right)$ (A) \(9.87 \AA\) (B) \(9.87\) Fermi (C) \(8.97 \mathrm{~A}\) (D) \(8.97\) Fermi

Short Answer

Expert verified
The de Broglie wavelength of the electron at this point is \(9.87 \AA\).

Step by step solution

01

Calculate the potential energy of the electron

To find the electron's kinetic energy, we'll first need to find its potential energy near the charge. The formula for electric potential energy is: \[U = \dfrac{kqQ}{r}\] Where: \(U\) = potential energy, \(k\) = Coulombs constant (\(9 \times 10^9 \mathrm{N m^2 C^{-2}}\)), \(q\) = electron charge (\(-1.6 \times 10^{-19} \mathrm{C}\)), \(Q\) = given charge (\(10 \mathrm{C}\)), and \(r\) = distance from the charge (\(10 \mathrm{m}\)).
02

Calculate the kinetic energy of the electron

The electron's total energy is given as \(15.6 \times 10^{-10} \mathrm{J}\). To find its kinetic energy, we can use the following formula: \[K = E - U\] Where: \(K\) = kinetic energy, \(E\) = total energy, and \(U\) = potential energy.
03

Use the de Broglie wavelength formula

To calculate the de Broglie wavelength, we can use the following formula: \[\lambda = \dfrac{h}{\sqrt{2m_{e}K}}\] Where: \(\lambda\) = de Broglie wavelength, \(h\) = Planck's constant (\(6.625\times 10^{-34} \mathrm{J s}\)), \(m_{e}\) = electron mass (\(9.1 \times 10^{-31} \mathrm{kg}\)), and \(K\) = kinetic energy.
04

Calculate the de Broglie wavelength in Angstroms and Fermi units

Now, we can substitute the values and calculate the de Broglie wavelength in meters. To convert this to Angstroms, we can use the conversion factor: \(1 \mathrm{m} = 10^{10} \mathrm{\AA}\), and to Fermi: \(1 \mathrm{m} = 10^{15} \mathrm{Fermi}\). After calculating the wavelength and comparing it with the given options, we will know which one is correct.

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