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Chapter 6: Problem 88

An electron is accelerated through an electric potential to a kinetic energy of \(18.6 \mathrm{keV}\). What is its characteristic wavelength? [Hint: Recall that the kinetic energy of a moving object is \(E=\frac{1}{2} m v^{2},\) where \(m\) is the mass of the object and \(\nu\) is the speed of the object.]

Short Answer

Expert verified
The characteristic wavelength of the electron is approximately \(2.428 \times 10^{-11}\) meters.

Step by step solution

01

Convert kinetic energy to SI units

The given kinetic energy is in keV, so first, we have to convert it into SI units (Joules). To do this, we use the conversion factor 1 eV = 1.6 x 10^{-19} J. \(18.6 \,\text{keV} = 18.6 \times 10^3 \,\text{eV}\) So, E = 18.6 x 10^3 eV x (1.6 x 10^{-19} J/eV) = 2.976 x 10^{-15} J
02

Use the given relation E = (1/2)mv^2

Now, we have the electron's kinetic energy in Joules. Next, we'll use the given relation E = (1/2)mv^2 and rearrange it to solve for the velocity (v). The mass of an electron (m) is 9.11 x 10^{-31} kg. \(v = \sqrt{\frac{2E}{m}} = \sqrt{\frac{2 \times 2.976 \times 10^{-15}\,\text{J}}{9.11 \times 10^{-31}\,\text{kg}}}\) Calculating the value for v, we get: \(v = 2.586 \times 10^7\,\text{m/s}\)
03

Apply the de Broglie wavelength formula

Now that we have the velocity of the electron, we can use the de Broglie wavelength formula to find its characteristic wavelength. The de Broglie wavelength formula is given as: \(\lambda = \frac{h}{m v}\) where h is the Planck's constant (6.626 x 10^{-34} Js), m is the mass of the electron (9.11 x 10^{-31} kg), and v is the velocity we calculated in step 2. Plugging in the values, we get: \(\lambda = \frac{6.626 \times 10^{-34}\,\text{Js}}{(9.11 \times 10^{-31}\,\text{kg}) \times (2.586 \times 10^7\,\text{m/s})}\) Calculating the value for λ, we get: \(\lambda = 2.428 \times 10^{-11}\,\text{m}\) Hence, the characteristic wavelength of the electron is approximately \(2.428 \times 10^{-11}\) meters.

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Most popular questions from this chapter

As shown in the accompanying photograph, an electric stove burner on its highest setting exhibits an orange glow. (a) When the burner setting is changed to low, the burner continues to produce heat but the orange glow disappears. How can this observation be explained with reference to one of the fundamental observations that led to the notion of quanta? (b) Suppose that the energy provided to the burner could be increased beyond the highest setting of the stove. What would we expect to observe with regard to visible light emitted by the burner? [Section 6.2\(]\)

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