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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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Most popular questions from this chapter

Wavelength \(\lambda_{\mathrm{A}}\) and \(\lambda_{\mathrm{B}}\) are incident on two identical metal plates and photo electrons are emitted. If \(\lambda_{\mathrm{A}}=2 \lambda_{\mathrm{B}}\), the maximum kinetic energy of photo electrons is \(\ldots \ldots \ldots\) (A) \(2 \mathrm{~K}_{\mathrm{A}}=\mathrm{K}_{\mathrm{B}}\) (B) \(\mathrm{K}_{\mathrm{A}}<\left(\mathrm{K}_{\mathrm{B}} / 2\right)\) (C) \(\mathrm{K}_{\mathrm{A}}=2 \mathrm{~K}_{\mathrm{B}}\) (D) \(\mathrm{K}_{\mathrm{A}}>\left(\mathrm{K}_{\mathrm{B}} / 2\right)\)

Work function of metal is \(2.5 \mathrm{eV}\) If wave length of light incident on metal plate is \(3000 \AA\), stopping potential of emitted electron will be....... $\left\\{\mathrm{h}=6.62 \times 10^{-34} \mathrm{~J} . \mathrm{s}, \mathrm{c}=3 \times 10^{8}(\mathrm{~m} / \mathrm{s})\right\\}$ (A) \(0.82 \mathrm{~V}\) (B) \(0.41 \mathrm{~V}\) (C) \(1.64 \mathrm{~V}\) (D) \(3.28 \mathrm{~V}\)

If kinetic energy of free electron is made double, change in de-Broglie wavelength will be........... (A) \(\sqrt{2}\) (B) \((1 / \sqrt{2})\) (C) 2 (D) \((1 / 2)\)

Suppose \(\Psi(\mathrm{x}, \mathrm{y}, \mathrm{z})\) represents a particle in three dimensional space, then probability of finding the particle in the unit volume at a given point \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) is $\ldots \ldots$ (A) inversely proportional to $\Psi^{\prime}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ (B) directly proportional \(\Psi^{*}\) (C) directly proportional to \(\mid \Psi \Psi^{*}\) (D) inversely proportional to \(\left|\Psi \Psi^{*}\right|\)

According to Einstein's photoelectric equation, graph of kinetic energy of emitted photo electrons from metal versus frequency of incident radiation is linear. Its slope.......... (A) depends on type of metal used (B) depends on intensity of radiation (C) depends on both metal used and intensity of radiation. (D) is same for all metals and free from intensity of radiation.
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