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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

A proton falls freely under gravity of Earth. Its de Broglie wavelength after \(10 \mathrm{~s}\) of its motion is \(\ldots \ldots \ldots\). Neglect the forces other than gravitational force. $\left[\mathrm{g}=10\left(\mathrm{~m} / \mathrm{s}^{2}\right), \mathrm{m}_{\mathrm{p}}=1.6 \times 10^{-27} \mathrm{~kg}, \mathrm{~h}=6.625 \times 10^{-34} \mathrm{~J}_{. \mathrm{s}}\right]$ (A) \(3.96 \AA\) (B) \(39.6 \AA\) (C) \(6.93 \AA\) (D) \(69.3 \AA\)

Wavelength of light incident on a photo-sensitive surface is reduced form \(3500 \AA\) to \(290 \mathrm{~mm}\). The change in stopping potential is $\ldots \ldots . .\left(\mathrm{h}=6.625 \times 10^{-24} \mathrm{~J} . \mathrm{s}\right)$ (A) \(42.73 \times 10^{-2} \mathrm{~V}\) (B) \(27.34 \times 10^{-2} \mathrm{~V}\) (C) \(73.42 \times 10^{-2} \mathrm{~V}\) (D) \(43.27 \times 10^{-2} \mathrm{~V}\)

Direction Read the following question choose if: (a) Both Assertion and Reason are true and Reason is correct explanation of Assertion. (b) Both Assertion and Reason are true, but Reason is not correct explanation of Assertion. (c) Assertion is true but the Reason is false. (d) Both Assertion and Reason is false. Assertion: Photo-electric effect can take place only when frequency is greater than threshold frequency \(\left(f_{0}\right)\) Reason: Electron is Fermions and photon is boson. (A) a (B) \(b\) (C) \(\mathrm{c}\) (D) d

If de-Broglie wavelength of electron is increased by \(1 \%\) its momentum \(\ldots \ldots\) (A) increases by \(1 \%\) (B) decreases by \(1 \%\) (C) increased by \(2 \%\) (D) decreases by \(2 \%\)

Read the paragraph carefully and select the proper choice from given multiple choices. According to Einstein when a photon of light of frequency for wavelength \(\lambda\) is incident on a photo sensitive metal surface of work function \(\Phi\). Where \(\Phi<\mathrm{hf}\) (here \(\mathrm{h}\) is Plank's constant) then the emission of photo-electrons place takes place. The maximum K.E. of emitted photo electrons is given by $\mathrm{K}_{\max }=\mathrm{hf}-\Phi .\( If the there hold frequency of metal is \)\mathrm{f}_{0}$ then \(\mathrm{hf}_{0}=\Phi\) (i) A metal of work function \(3.3 \mathrm{eV}\) is illuminated by light of wave length \(300 \mathrm{~nm}\). The maximum \(\mathrm{K} . \mathrm{E}>\) of photo- electrons is $=\ldots \ldots \ldots . \mathrm{eV} .\left(\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J} \cdot \mathrm{sec}\right)$ (A) \(0.825\) (B) \(0.413\) (C) \(1.32\) (D) \(1.65\) (ii) Stopping potential of emitted photo-electron is $=\ldots \ldots . \mathrm{V}$. (A) \(0.413\) (B) \(0.825\) (C) \(1.32\) (D) \(1.65\) (iii) The threshold frequency fo $=\ldots \ldots \ldots \times 10^{14} \mathrm{~Hz}$. (A) \(4.0\) (B) \(4.2\) (C) \(8.0\) (D) \(8.4\)

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