College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Recoil Velocity
The amount of backward thrust or force experienced by a person when he/she shoots a gun in the forward direction is called recoil velocity. This phenomenon always follows the law of conservation of linear momentum.
Question
![**Question:**
What is the minimum x-ray wavelength produced for a dental x-ray machine operated at 30 kV?
**Explanation:**
To determine the minimum x-ray wavelength, we use the energy-wavelength relationship in the context of x-ray production. The energy \(E\) of the x-ray photon is related to its wavelength \(\lambda\) by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \(E\) is the energy of the x-ray photon,
- \(h\) is Planck’s constant (\(6.626 \times 10^{-34}\) Js),
- \(c\) is the speed of light in a vacuum (\(3 \times 10^8\) m/s),
- \(\lambda\) is the wavelength of the x-ray.
For a dental x-ray machine operated at 30 kV (kilovolts), the maximum energy of the produced x-rays is 30 keV (kilo-electron volts). To convert this energy into joules (J), we use the conversion factor \(1 \text{eV} = 1.602 \times 10^{-19} \text{ J}\):
\[ 30 \text{ keV} = 30,000 \times 1.602 \times 10^{-19} \text{ J} \approx 4.806 \times 10^{-15} \text{ J}\]
Now, we can rearrange the energy-wavelength equation to solve for \(\lambda\):
\[ \lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \text{ Js} \times 3 \times 10^8 \text{ m/s}}{4.806 \times 10^{-15} \text{ J}} \approx 2.59 \times 10^{-11} \text{ m} \]
Thus, the minimum x-ray wavelength produced for a dental x-ray machine operated at 30 kV is approximately \(2.59 \times 10^{-11}\) meters, or 0.0259 nanometers (nm).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F093e7116-3e01-49ad-9157-7a22a5caeb16%2F03ca9deb-65e6-435f-93d8-0ca72265a981%2F160imif_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
What is the minimum x-ray wavelength produced for a dental x-ray machine operated at 30 kV?
**Explanation:**
To determine the minimum x-ray wavelength, we use the energy-wavelength relationship in the context of x-ray production. The energy \(E\) of the x-ray photon is related to its wavelength \(\lambda\) by the equation:
\[ E = \frac{hc}{\lambda} \]
where:
- \(E\) is the energy of the x-ray photon,
- \(h\) is Planck’s constant (\(6.626 \times 10^{-34}\) Js),
- \(c\) is the speed of light in a vacuum (\(3 \times 10^8\) m/s),
- \(\lambda\) is the wavelength of the x-ray.
For a dental x-ray machine operated at 30 kV (kilovolts), the maximum energy of the produced x-rays is 30 keV (kilo-electron volts). To convert this energy into joules (J), we use the conversion factor \(1 \text{eV} = 1.602 \times 10^{-19} \text{ J}\):
\[ 30 \text{ keV} = 30,000 \times 1.602 \times 10^{-19} \text{ J} \approx 4.806 \times 10^{-15} \text{ J}\]
Now, we can rearrange the energy-wavelength equation to solve for \(\lambda\):
\[ \lambda = \frac{hc}{E} = \frac{6.626 \times 10^{-34} \text{ Js} \times 3 \times 10^8 \text{ m/s}}{4.806 \times 10^{-15} \text{ J}} \approx 2.59 \times 10^{-11} \text{ m} \]
Thus, the minimum x-ray wavelength produced for a dental x-ray machine operated at 30 kV is approximately \(2.59 \times 10^{-11}\) meters, or 0.0259 nanometers (nm).
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