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Chapter 36: Problem 69

What is the wavelength of an electron that is accelerated from rest through a potential difference of \(1.00 \cdot 10^{-5} \mathrm{~V} ?\)

Short Answer

Expert verified
Given an electron accelerated from rest through a potential difference of \(1.00 \cdot 10^{-5} \mathrm{~V}\), please calculate its wavelength. The wavelength of the electron, after being accelerated by a potential difference of \(1.00 \cdot 10^{-5} \mathrm{~V}\), would be approximately \(6.15 \cdot 10^{-7} \mathrm{m}\).

Step by step solution

01

Find the final kinetic energy

We have to find the final kinetic energy of the electron, denoted as \(K_f\), by using the formula \(K_f = K_0 + qV\). Since the electron is initially at rest \(K_0=0\), we have: \(K_f = 0 + qV\) where \(q\) = the charge of the electron, \(1.60 \cdot 10^{-19} \mathrm{C}\), and \(V\) = the potential difference, \(1.00 \cdot 10^{-5} \mathrm{V}\). Then, calculate the final kinetic energy: \(K_f = (1.60 \cdot 10^{-19}\ \mathrm{C})(1.00 \cdot 10^{-5}\ \mathrm{V})\) \(K_f = 1.60 \cdot 10^{-24}\ \mathrm{J}\).
02

Find the momentum of the electron

We can find the momentum of the electron by using the relativistic energy-momentum relationship: \(E^2 = (pc)^2 + (mc^2)^2\) Given that \(E = \sqrt{K_f^2+(mc^2)^2}\) and \(E = pc\), we can calculate \(p\): \(p = \dfrac{\sqrt{K_f^2 + (mc^2)^2}}{c}\) where \(K_f = 1.60 \cdot 10^{-24}\) J, \(c\) = speed of light, \(3.00 \cdot 10^8\) m/s, and \(m\) = mass of the electron, \(9.11 \cdot 10^{-31}\) kg. Solve for \(p\): \(p = \dfrac{\sqrt{(1.60 \cdot 10^{-24})^2 + (9.11 \cdot 10^{-31} \cdot (3.00 \cdot 10^8)^2)^2}}{3.00 \cdot 10^8}\) \(p \approx 1.077 \cdot 10^{-28} \ \mathrm{kg\ m/s}\).
03

Calculate the wavelength of the electron

Lastly, we use the de Broglie wavelength formula \(\lambda = \dfrac{h}{p}\) to find the wavelength of the electron: \(\lambda = \dfrac{h}{p}\) where \(h\) = Planck's constant, \(6.626 \cdot 10^{-34} \mathrm{Js}\), and \(p \approx 1.077 \cdot 10^{-28} \ \mathrm{kg\ m/s}\). Solving for \(\lambda\): \(\lambda = \dfrac{6.626 \cdot 10^{-34} \mathrm{Js}}{1.077 \cdot 10^{-28} \ \mathrm{kg\ m/s}}\) \(\lambda \approx 6.15 \cdot 10^{-7} \mathrm{m}\). Thus, the wavelength of the electron accelerated from rest through a potential difference of \(1.00 \cdot 10^{-5} \mathrm{~V}\) is approximately \(6.15 \cdot 10^{-7} \mathrm{m}\).

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