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Chapter 27: Problem 20

Show that the cutoff frequency for an x-ray tube is proportional to the potential difference through which the electrons are accelerated.

Short Answer

Expert verified
Question: Show that the cutoff frequency for an x-ray tube is proportional to the potential difference through which the electrons are accelerated. Answer: The cutoff frequency (f_max) for an x-ray tube is directly proportional to the potential difference (V) through which the electrons are accelerated, as shown by the equation f_max = (eV)/h, where e is the elementary charge and h is Planck's constant.

Step by step solution

01

Write down the energy of an electron in terms of potential difference

Use the formula for the energy of an electron when it is accelerated through a potential difference V: E = eV where E is the energy of the electron, e is the elementary charge (1.6 x 10^{-19} C), and V is the potential difference through which the electron is accelerated.
02

Calculate the maximum frequency of the x-rays produced

The maximum frequency (cutoff frequency) of the x-rays produced can be found using Planck's equation: E = hf_{max} where h is Planck's constant (6.63 x 10^{-34} Js) and f_{max} is the maximum frequency of the x-rays produced.
03

Find the relationship between theelectron energy and maximum frequency

Combine the two equations from Steps 1 and 2 to find the relationship between electron energy and maximum frequency: eV = hf_{max}
04

Express the maximum frequency in terms of potential difference

Solve the equation from Step 3 for the maximum frequency (cutoff frequency): f_{max} = \frac{eV}{h}
05

Conclude the proportionality

The equation from Step 4 shows that the cutoff frequency (f_{max}) is directly proportional to the potential difference (V) through which the electrons are accelerated. Therefore, we have shown that the cutoff frequency for an x-ray tube is proportional to the potential difference through which the electrons are accelerated.

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Most popular questions from this chapter

A photon of wavelength \(0.14800 \mathrm{nm}\), traveling due east, is scattered by an electron initially at rest. The wavelength of the scattered photon is \(0.14900 \mathrm{nm}\) and it moves at an angle \(\theta\) north of east. (a) Find \(\theta .\) (b) What is the south component of the electron's momentum?
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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.
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