Chapter 4: Problem 109
An X-ray tube is operated at 50,000 volts. The short-wavelength limit of the X-rays produced is (a) \(0.1245 \AA\) (b) \(0.3485 \mathrm{~A}\) (c) \(0.2485 \mathrm{~A}\) (d) \(0.03456 \AA\)
Short Answer
Expert verified
Option (c) \(0.2485 \; \text{Ã…}\) is the correct answer.
Step by step solution
01
Understanding the Problem
The problem asks us to find the short-wavelength limit of X-rays produced when operating an X-ray tube at a given voltage. This involves understanding the relationship between photon energy and wavelength.
02
Using the Relevant Formula
The energy of an X-ray photon at the short-wavelength limit can be calculated using the equation \( E = hf = \frac{hc}{\lambda} \), where \( \lambda \) is the wavelength, \( h \) is Planck's constant \( 6.626 \times 10^{-34} \; \text{Js} \), \( c \) is the speed of light \( 3.00 \times 10^8 \; \text{m/s} \). For the short-wavelength limit, the energy \( E \) is also the maximum energy provided by the tube, which is \( E = eV \), where \( e \) is the elementary charge \( 1.602 \times 10^{-19} \; \text{C} \) and \( V \) is the voltage.
03
Calculating the Wavelength
Substitute \( V = 50,000 \) volts and \( e = 1.602 \times 10^{-19} \; \text{C} \) into the equation \( E = eV = \frac{hc}{\lambda} \). Therefore, \( \lambda = \frac{hc}{eV} \). Using the known values, we find \[ \lambda = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^8}{1.602 \times 10^{-19} \times 50,000} \] which calculates to approximately \( \lambda = 0.2485 \; \text{Ã…} \).
04
Conclusion and Answer Selection
The calculated wavelength is \( 0.2485 \; \text{Ã…}\). Therefore, the option that closely matches is option (c) \(0.2485 \; \text{Ã…}\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Short-wavelength limit
In X-ray production, the **short-wavelength limit** is a term used to describe the shortest possible wavelength an X-ray photon can have while being generated by an X-ray tube. This term is essential because it directly indicates the maximum energy of X-rays that a specific tube can produce. By understanding this, you can determine the potential applications of the X-rays generated, such as medical imaging or material analysis.
The short-wavelength limit is related to the maximum photon energy that can be achieved in the X-ray tube. If you integrate the concept of photon energy with wavelength, you realize that shorter wavelengths correspond to higher energy photons. This is why the short-wavelength limit is important: it tells us how energetic and therefore how penetrating the produced X-rays can be.
In practical terms, knowing the short-wavelength limit helps in selecting the correct type of X-ray tube for specific tasks, ensuring enough energy for imaging or penetration of materials.
The short-wavelength limit is related to the maximum photon energy that can be achieved in the X-ray tube. If you integrate the concept of photon energy with wavelength, you realize that shorter wavelengths correspond to higher energy photons. This is why the short-wavelength limit is important: it tells us how energetic and therefore how penetrating the produced X-rays can be.
In practical terms, knowing the short-wavelength limit helps in selecting the correct type of X-ray tube for specific tasks, ensuring enough energy for imaging or penetration of materials.
Photon energy
The energy of a photon is a fundamental concept in understanding the behavior of electromagnetic radiation like X-rays. **Photon energy** is calculated by the formula: \[ E = hf = \frac{hc}{\lambda} \] where:
This high energy is what makes X-rays capable of penetrating materials and is a reason for their use in imaging and medical diagnostics. The ability to calculate photon energy precisely is important in predicting X-ray behavior and safely using them in various applications.
- \( E \) is the energy of the photon.
- \( h \) is Planck's constant \( (6.626 \times 10^{-34} \text{ Js}) \).
- \( c \) is the speed of light \( (3.00 \times 10^8 \text{ m/s}) \).
- \( \lambda \) is the wavelength of the photon.
This high energy is what makes X-rays capable of penetrating materials and is a reason for their use in imaging and medical diagnostics. The ability to calculate photon energy precisely is important in predicting X-ray behavior and safely using them in various applications.
Planck's constant
**Planck's constant** \( (h) \) is a pivotal constant in physics with a value of approximately \( 6.626 \times 10^{-34} \text{ Js} \). It plays a critical role in quantum mechanics and the calculation of photon energy. Planck's constant serves as a fundamental link between the frequency of electromagnetic radiation and its energy.
When you observe the formula \( E = hf \), you notice that Planck's constant helps determine how much energy corresponds to a specific frequency of electromagnetic waves, like those of X-rays. This energy-frequency relationship is crucial because it forms the basis of how energy levels are defined in quantum mechanics and how different types of radiation interact with matter.
In X-ray production, Planck's constant is essential to calculate and understand the extent of energy that can be harnessed or produced, offering insights into the type of diagnostic or therapeutic applications X-rays may serve.
When you observe the formula \( E = hf \), you notice that Planck's constant helps determine how much energy corresponds to a specific frequency of electromagnetic waves, like those of X-rays. This energy-frequency relationship is crucial because it forms the basis of how energy levels are defined in quantum mechanics and how different types of radiation interact with matter.
In X-ray production, Planck's constant is essential to calculate and understand the extent of energy that can be harnessed or produced, offering insights into the type of diagnostic or therapeutic applications X-rays may serve.
Voltage in X-ray production
The role of **voltage in X-ray production** can't be overstated. It essentially determines the maximum kinetic energy that electrons can achieve before they strike the target material to produce X-rays. The voltage applied in the X-ray tube governs the energy of the resulting X-ray photons through the relation:\[ E = eV \] where:
Understanding how voltage affects X-ray production is vital for optimizing the performance of X-ray equipment. It ensures the produced X-rays are suitable for their intended purpose while minimizing unnecessary exposure to radiation.
- \( E \) is the energy of the X-ray photon.
- \( e \) is the elementary charge \( (1.602 \times 10^{-19} \text{ C}) \).
- \( V \) is the applied voltage across the tube.
Understanding how voltage affects X-ray production is vital for optimizing the performance of X-ray equipment. It ensures the produced X-rays are suitable for their intended purpose while minimizing unnecessary exposure to radiation.