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Chapter 17: Problem 2419

An electron moving with velocity \(0.6 \mathrm{c}\), then de-brogly wavelength associated with is \(\ldots \ldots \ldots\) (rest mars of electron, \(\mathrm{m}_{0}=9.1 \times 10^{-31}(\mathrm{k} / \mathrm{s})\) \(\mathrm{h}=6.63 \times 10^{-34} \mathrm{Js}\) (A) \(3.24 \times 10^{-12} \mathrm{~m}\) (B) \(32.4 \times 10^{-12} \mathrm{~m}\) (C) \(320 \times 10^{-12} \mathrm{~m}\) (D) \(3.29 \times 10^{-14} \mathrm{~m}\)

Short Answer

Expert verified
The de Broglie wavelength associated with the electron moving with velocity \(0.6c\) is approximately \(3.24 \times 10^{-12} \, m\).

Step by step solution

01

Identify the de Broglie wavelength formula

The de Broglie wavelength formula is given by: \[ \lambda = \frac{h}{p} \] where \(\lambda\) is the de Broglie wavelength, \(h\) is the Planck's constant, and \(p\) is the momentum of the particle.
02

Calculate the electron's momentum

The momentum formula for a relativistic particle is given by: \[ p = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}} \] Let's plug the values of mass of electron (\(m_0 = 9.1 \times 10^{-31} \, kg\)), velocity (\(v = 0.6c\)), and the speed of light (\(c = 3 \times 10^8 \, m/s\)) into the formula. \[ p = \frac{9.1 \times 10^{-31} \times 0.6 \times (3 \times 10^8)}{\sqrt{1 - 0.6^2}} \]
03

Solve for the momentum

After simplifying and calculating the denominator, we find: \[ p = \frac{9.1 \times 10^{-31} \times 0.6 \times (3 \times 10^8)}{\sqrt{1 - 0.36}} = \frac{9.1 \times 10^{-31} \times 0.6 \times (3 \times 10^8)}{\sqrt{0.64}} = \frac{9.1 \times 10^{-31} \times 0.6 \times (3 \times 10^8)}{0.8} \] Further calculation gives: \[ p \approx 2.051 \times 10^{-22} \, kg \cdot m/s \]
04

Calculate de Broglie wavelength

Now, we'll substitute the values of Planck's constant (\(h = 6.63 \times 10^{-34} \, Js\)) and the calculated momentum into the de Broglie wavelength formula: \[ \lambda = \frac{6.63 \times 10^{-34}}{2.051 \times 10^{-22}} \approx 3.23 \times 10^{-12} \, m \]
05

Match the calculated wavelength with the given options

Comparing our calculated wavelength with the given options, it's closest to option (A): \[ \lambda = 3.24 \times 10^{-12} \, m \]

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Most popular questions from this chapter

An electron is accelerated between two points having potential $20 \mathrm{~V}\( and \)40 \mathrm{~V}$, de-Broglic wavelength of electron is \(\ldots \ldots\) (A) \(0.75 \AA\) (B) \(7.5 \AA\) (C) \(2.75 \AA\) (D) \(0.75 \mathrm{~nm}\)

Kinetic energy of proton accelerated under p.d. \(1 \mathrm{~V}\) will be........ (A) \(1840 \mathrm{eV}\) (B) \(13.6 \mathrm{eV}\) (C) \(1 \mathrm{eV}\) (D) \(0.54 \mathrm{eV}\)

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If we take accelerating voltage \(\mathrm{V}=50 \mathrm{~V}\), electric charge of electron \(\mathrm{e}=1.6 \times 10^{-19} \mathrm{C}\) and mass of electron \(\mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}\) find the wavelength of concerned election. (A) \(0.1735 \mathrm{~A}\) (B) \(1.735 \AA\) (C) \(17.35 \AA\) (D) \(1735 \AA\)

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