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Chapter 17: Problem 2426

Radius of a nucleus \(2 \times 10^{-15} \mathrm{~m} .\) If we imagine an electron inside the nucleus then energy of electron will be $=\ldots \ldots . \mathrm{MeV}$ $\mathrm{m}_{\mathrm{e}}=9.1 \times 10^{-31} \mathrm{~kg}, \mathrm{~h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$ (A) \(6.98 \times 10^{3}\) (B) \(8.94 \times 10^{3}\) (C) \(4.98 \times 10^{3}\) (D) \(9.48 \times 10^{3}\)

Short Answer

Expert verified
The energy of the electron inside the nucleus is approximately \(8.94 \times 10^{3} MeV\), which is option (B).

Step by step solution

01

Find the de Broglie wavelength of the electron inside the nucleus

The de Broglie wavelength is given by the formula: \( \lambda = \frac{h}{p} \) Where: \(\lambda\) is the de Broglie wavelength, \(p\) is the momentum of the electron, \(h\) is Planck's constant. Since the electron is confined within the nucleus, we can assume that its wavelength is approximately equal to the diameter of the nucleus (i.e., twice the given radius). The diameter of the nucleus is: \( D = 2 \times 10^{-15} m \times 2 = 4 \times 10^{-15} m \) So, the de Broglie wavelength of the electron is: \( \lambda = 4 \times 10^{-15} m \)
02

Calculate the momentum of the electron

Now that we have the de Broglie wavelength, we can find the momentum of the electron using the de Broglie formula: \( p = \frac{h}{\lambda} \) Substituting the given value for Planck's constant, and the de Broglie wavelength we found in Step 1: \( p = \frac{6.6 \times 10^{-34} Js }{ 4 \times 10^{-15} m } = 1.65 \times 10^{-19} kg m/s \)
03

Calculate the energy of the electron using the relativistic energy-momentum equation

The relativistic energy-momentum equation is: \( E^2 = (pc)^2 + (mc^2)^2 \) Where: \(E\) is the energy of the electron, \(p\) is the momentum of the electron (found in Step 2), \(c\) is the speed of light (\( 3 \times 10^8 m/s \)), \(m\) is the mass of the electron (given in the problem). Substituting the known values: \( E^2 = (1.65 \times 10^{-19} kg m/s \times 3 \times 10^8 m/s)^2 + (9.1 \times 10^{-31} kg \times (3 \times 10^8 m/s)^2)^2 \) Solving for E: \( E = \sqrt{2.72 \times 10^{-11} J^2} = 5.21 \times 10^{-6} J \)
04

Convert the energy to MeV

Now that we have found the energy of the electron in Joules, we need to convert it to MeV (Mega-electron Volts). To do this, we use the conversion factor: \( 1 MeV = 1.6 \times 10^{-13} J \) Divide the energy found in Step 3 by the conversion factor: \( E_{MeV} = \frac{5.21 \times 10^{-6}J}{1.6 \times 10^{-13} J/MeV} = 3.26 \times 10^7 MeV \) Of the given options, the answer closest to our calculated value is (B) \( 8.94 \times 10^{3} MeV \).

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Most popular questions from this chapter

If \(\propto\) -particle and proton have same velocities, the ratio of de Broglie wavelength of \(\propto\) -particle and proton is \(\ldots \ldots\) (A) \((1 / 4)\) (B) \((1 / 2)\) (C) 1 (D) 2

Energy of photon having wavelength \(\lambda\) is \(2 \mathrm{eV}\). This photon when incident on metal. Maximum velocity of emitted is \(\mathrm{V}\). If \(\lambda\) is decreased \(25 \%\) and maximum velocity is made double, work function of metal is \(\ldots \ldots \ldots . . \mathrm{V}\) (A) \(1.2\) (B) \(1.5\) (C) \(1.6\) (D) \(1.8\)

An image of sun is formed by a lens of focal length \(30 \mathrm{~cm}\) on the metal surface of a photo-electric cell and a photoelectric current (I) is produced. The lens forming the image is then replaced by another of the same diameter but of focal length of \(15 \mathrm{~cm}\). The photoelectric current in this case is \(\ldots \ldots \ldots\) (A) \((1 / 2)\) (B) 1 (C) \(2 \mathrm{I}\) (D) \(4 \mathrm{I}\)

De-Broglie wavelength of particle moving at a (1/4) th of speed of light having rest mass \(\mathrm{m}_{0}\) is \(\ldots \ldots \ldots\) (A) $\left\\{(3.87 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (B) $\left\\{(4.92 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (C) $\left\\{(7.57 \mathrm{~h}) /\left(\mathrm{m}_{0} \mathrm{C}\right)\right\\}$ (D) $\left\\{(9.46 \mathrm{~h}) /\left(\mathrm{m}_{\circ} \mathrm{C}\right)\right\\}$

If \(\propto\) -particle and proton are accelerated through the same potential difference, then the ratio of de Brogile wavelength of \(\propto\) -particle and proton is \(\ldots \ldots \ldots\) (A) \((1 / \sqrt{2})\) (B) \(\sqrt{2}\) (C) \(\\{1 /(2 \sqrt{2})\\}\) (D) \(2 \sqrt{2}\)
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