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Chapter 27: Problem 18

The potential difference in an x-ray tube is \(40.0 \mathrm{kV}\) What is the minimum wavelength of the continuous x-ray spectrum emitted from the tube?

Short Answer

Expert verified
Answer: The minimum wavelength of the continuous x-ray spectrum emitted from the tube is 3.10 x 10^-11 m.

Step by step solution

01

Convert potential difference to energy

First, we need to convert the potential difference in the x-ray tube from kilovolts (kV) to joules (J) to find the maximum energy of the emitted x-ray, using the formula: \(E = qV\), where \(E\) is the energy, \(q\) is the charge of an electron (\(1.6 \times 10^{-19} \:\mathrm{C}\)), and \(V\) is the potential difference. In this case, \(V = 40.0 \times 10^{3} \:\mathrm{V}\).
02

Calculate the maximum energy x-ray

Now we can calculate the maximum energy x-ray produced in the tube by multiplying the charge of an electron with the potential difference: \(E = (1.6 \times 10^{-19} \:\mathrm{C})(40.0 \times 10^{3} \:\mathrm{V}) = 6.4 \times 10^{-15} \:\mathrm{J}\).
03

Relate energy to wavelength using Planck's equation

Planck's equation relates the energy of a photon to its wavelength: \(E = \dfrac{hc}{\lambda}\), where \(E\) is the energy, \(h\) is Planck's constant (\(6.626 \times 10^{-34}\: \mathrm{J\:s}\)), \(c\) is the speed of light (\(3.0 \times 10^{8}\: \mathrm{m/s}\)), and \(\lambda\) is the wavelength. Since we want the minimum wavelength, we should use the maximum energy obtained in Step 2.
04

Solve for the minimum wavelength

By rearranging Planck's equation to isolate \(\lambda\), we get: \(\lambda = \dfrac{hc}{E}\). Plugging in the values, we calculate the minimum wavelength: \(\lambda = \dfrac{(6.626 \times 10^{-34}\: \mathrm{J\:s})(3.0 \times 10^{8}\: \mathrm{m/s})}{6.4 \times 10^{-15} \:\mathrm{J}} = 3.10 \times 10^{-11} \:\mathrm{m}\). Hence, the minimum wavelength of the continuous x-ray spectrum emitted from the tube is \(3.10 \times 10^{-11} \:\mathrm{m}\).

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Most popular questions from this chapter

The Paschen series in the hydrogen emission spectrum is formed by electron transitions from \(n_{\mathrm{i}}>3\) to \(n_{\mathrm{f}}=3.\) (a) What is the longest wavelength in the Paschen series? (b) What is the wavelength of the series limit (the lower bound of the wavelengths in the series)? (c) In what part or parts of the EM spectrum is the Paschen series found (IR, visible, UV, etc.)?
A positron and an electron that were at rest suddenly vanish and two photons of identical frequency appear. What is the wavelength of each of these photons?
An incident photon of wavelength \(0.0100 \mathrm{nm}\) is Compton scattered; the scattered photon has a wavelength of \(0.0124 \mathrm{nm} .\) What is the change in kinetic energy of the electron that scattered the photon?
X-rays of wavelength \(10.0 \mathrm{pm}\) are incident on a target. Find the wavelengths of the \(x\) -rays scattered at (a) \(45.0^{\circ}\) and (b) \(90.0^{\circ} .\) (tutorial: Compton scattering)
Suppose that you have a glass tube filled with atomic hydrogen gas (H, not \(\mathrm{H}_{2}\) ). Assume that the atoms start out in their ground states. You illuminate the gas with monochromatic light of various wavelengths, ranging through the entire IR, visible, and UV parts of the spectrum. At some wavelengths, visible light is emitted from the \(\mathrm{H}\) atoms. (a) If there are two and only two visible wavelengths in the emitted light, what is the wavelength of the incident radiation? (b) What is the largest wavelength of incident radiation that causes the \(\mathrm{H}\) atoms to emit visible light? What wavelength(s) is/are emitted for incident radiation at that wavelength? (c) For what wavelengths of incident light are hydrogen ions \(\left(\mathrm{H}^{+}\right)\) formed?
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