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Chapter 27: Problem 69

What potential difference must be applied to an x-ray tube to produce x-rays with a minimum wavelength of \(45.0 \mathrm{pm} ?\)

Short Answer

Expert verified
Answer: The required potential difference is approximately 2.94 x 10^4 V.

Step by step solution

01

Convert the given wavelength from pm to meters

First, we need to convert the given minimum wavelength from picometers to meters: \(\lambda_\text{min} = 45.0 \mathrm{pm} \times \dfrac{1 \mathrm{m}}{10^{12} \mathrm{pm}} = 45.0 \times 10^{-12} \mathrm{m}\)
02

Rearrange the Duane-Hunt law to solve for V

Next, we rearrange the Duane-Hunt law to isolate \(V\) on one side of the equation: \(V = \dfrac{hc}{e\lambda_\text{min}}\)
03

Substitute the known values and calculate V

Now, we substitute the known values of \(h\), \(c\), \(e\), and \(\lambda_\text{min}\) into the equation and calculate \(V\): \(V = \dfrac{(6.6261 \times 10^{-34} \mathrm{J\cdot s})(2.9979 \times 10^8 \mathrm{m/s})}{(1.6022 \times 10^{-19} \mathrm{C})(45.0 \times 10^{-12} \mathrm{m})}\) \(V \approx 2.94 \times 10^4 \mathrm{V}\) The required potential difference to produce x-rays with a minimum wavelength of \(45.0 \mathrm{pm}\) is approximately \(2.94 \times 10^4 \mathrm{V}\).

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Most popular questions from this chapter

An x-ray photon of initial frequency \(3.0 \times 10^{19} \mathrm{Hz}\) collides with a free electron at rest; the scattered photon moves off at \(90^{\circ} .\) What is the frequency of the scattered photon?
A hydrogen atom in its ground state is immersed in a continuous spectrum of ultraviolet light with wavelengths ranging from 96 nm to 110 nm. After absorbing a photon, the atom emits one or more photons to return to the ground state. (a) What wavelength(s) can be absorbed by the \(\mathrm{H}\) atom? (b) For each of the possibilities in (a), if the atom is at rest before absorbing the UV photon, what is its recoil speed after absorption (but before emitting any photons)? (c) For each of the possibilities in (a), how many different ways are there for the atom to return to the ground state?
Photons with a wavelength of 400 nm are incident on an unknown metal, and electrons are ejected from the metal. However, when photons with a wavelength of \(700 \mathrm{nm}\) are incident on the metal, no electrons are ejected. (a) Could this metal be cesium with a work function of \(1.8 \mathrm{eV} ?\) (b) Could this metal be tungsten with a work function of 4.6 eV? (c) Calculate the maximum kinetic energy of the ejected electrons for each possible metal when 200 -nm photons are incident on it.
The minimum energy required to remove an electron from a metal is 2.60 eV. What is the longest wavelength photon that can eject an electron from this metal?
Follow the steps outlined in this problem to estimate the time lag (predicted classically but not observed experimentally) in the photoelectric effect. Let the intensity of the incident radiation be $0.01 \mathrm{W} / \mathrm{m}^{2} .\( (a) If the area of the atom is \)(0.1 \mathrm{nm})^{2},$ find the energy per second falling on the atom. (b) If the work function is 2.0 eV, how long would it take (classically) for enough energy to fall on this area to liberate one photoelectron? (c) Explain briefly, using the photon model, why this time lag is not observed.
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