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Chapter 19: Problem 41

The speed of an electron having a wavelength of the order of \(1 \AA\) will be (A) \(7.25 \times 10^{6} \mathrm{~m} / \mathrm{s}\) (B) \(6.26 \times 10^{6} \mathrm{~m} / \mathrm{s}\) (C) \(5.25 \times 10^{6} \mathrm{~m} / \mathrm{s}\) (D) \(4.24 \times 10^{6} \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The speed of an electron having a wavelength of the order of \(1 \AA\) will be approximately \(7.25 \times 10^6 \, \text{m/s}\) (Option A).

Step by step solution

01

Write down the de Broglie wavelength formula

The de Broglie wavelength formula is given by: \(\lambda = \dfrac{h}{p}\) Where \(\lambda\) is the wavelength, \(h\) is the Planck constant (\(6.626 \times 10^{-34} \, \text{Js}\)), and \(p\) is the particle's momentum.
02

Write down the relationship between momentum and velocity

The momentum of a particle is given by the product of its mass and velocity: \(p = m_e v\) Where \(m_e\) is the mass of the electron (\(9.109 \times 10^{-31} \, \text{kg}\)) and \(v\) is its velocity.
03

Combine the two formulas to relate wavelength and velocity

We'll now eliminate momentum from the equations by substituting the equation for momentum into the de Broglie wavelength formula: \(\lambda = \dfrac{h}{m_e v}\)
04

Solve for velocity

To find the velocity of the electron, we'll rearrange the equation to solve for \(v\): \(v = \dfrac{h}{m_e \lambda}\)
05

Plug in the given values and compute the velocity

Substituting the given wavelength (\(\lambda = 1 \, \text{脜} = 1 \times 10^{-10} \, \text{m}\)), Planck constant (\(h = 6.626 \times 10^{-34} \, \text{Js}\)), and electron mass (\(m_e = 9.109 \times 10^{-31} \, \text{kg}\)), we get: \(v = \dfrac{6.626 \times 10^{-34}}{9.109 \times 10^{-31} \times 1 \times 10^{-10}}\) \(v \approx 7.27 \times 10^6 \,\text{m/s}\) Comparing this value to the provided options, we find that the closest answer is: (A) \(7.25 \times 10^6 \, \text{m/s}\)

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Most popular questions from this chapter

In a hydrogen atom, an electron of mass \(m\) and charge \(e\) is in an orbit of radius \(r\) making \(n\) revolutions per second. If the mass of the hydrogen nucleus is \(M\), the magnetic moment associated with the orbital motion of the electron is (A) \(\frac{\pi n e r^{2} m}{M}\) (B) \(\frac{\pi n e r^{2} M}{m}\) (C) \(\frac{\pi n e r^{2} m}{(M+m)}\) (D) \(\pi n e r^{2}\)

Electromagnetic radiation whose electric component varies with time as \(E=C_{1}\left(C_{2}+C_{3} \cos \omega t\right) \cos \omega_{0} t\), here \(C_{1}, C_{2}\) and \(C_{3}\) are constants, is incident on lithium and liberates photoelectrons. Let the kinetic energy of most energetic electrons be \(0.592 \times 10^{-19} \mathrm{~J}\). Given that \(\omega_{0}=3.6 \times 10^{15} \mathrm{rad} / \mathrm{s}\) and \(\omega=6 \times 10^{14}\) rad/sec. The work function of lithium is (take planks constant \(h=6.6 \times 10^{-34} \mathrm{MKS}\) ). (A) \(1.2 \mathrm{eV}\) (B) \(1.5 \mathrm{eV}\) (C) \(2.1 \mathrm{eV}\) (D) \(2.39 \mathrm{eV}\)

In hydrogen atom, if potential energy of electron in ground state is assumed to be zero, then its energy in first excited state is equal to (A) \(-3.4 \mathrm{eV}\) (B) \(23.8 \mathrm{eV}\) (C) \(17.2 \mathrm{eV}\) (D) \(-6.8 \mathrm{eV}\)

An electron of mass \(m\) and charge \(e\) is accelerated by a potential difference \(V .\) It then enters a uniform magnetic field \(B\) applied perpendicular to its path. The radius of the circular path of the electron is (A) \(r=\left(\frac{2 m V}{e B^{2}}\right)^{\frac{1}{2}}\) (B) \(r=\left(\frac{2 m e V}{B^{2}}\right)^{\frac{1}{2}}\) (C) \(r=\left(\frac{2 m B}{e V^{2}}\right)^{\frac{1}{2}}\) (D) \(r=\left(\frac{2 B^{2} V}{e m}\right)^{\frac{1}{2}}\)

The time by a photoelectron to come out after the photon strikes is approximately (A) \(10^{-1} \mathrm{~s}\) (B) \(10^{-4} \mathrm{~s}\) (C) \(10^{-10} \mathrm{~s}\) (D) \(10^{-16} \mathrm{~s}\)
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