An electron has a kinetic energy of 3.00 eV. Find its wavelength.
Related questions
Question
100%
![**Question:**
An electron has a kinetic energy of 3.00 eV. Find its wavelength.
**Explanation:**
To find the wavelength of an electron with a given kinetic energy, one can use the de Broglie wavelength formula. The de Broglie wavelength (\(\lambda\)) for a particle is given by:
\[\lambda = \frac{h}{p}\]
where
- \(h\) is the Planck constant (\(6.626 \times 10^{-34} \text{ Js}\)),
- \(p\) is the momentum of the particle.
First, we need to find the momentum (\(p\)). The kinetic energy (KE) of an electron can be related to its momentum using the following equation:
\[ KE = \frac{p^2}{2m} \]
where \(m\) is the mass of the electron (\(9.11 \times 10^{-31} \text{ kg}\)).
Given \(KE = 3.00 \text{ eV}\), we need to convert this energy into Joules (since the Planck constant is in Joules·second). The conversion factor is:
\[ 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \]
So, \(3.00 \text{ eV}\) is:
\[ 3.00 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} = 4.806 \times 10^{-19} \text{ J} \]
Using the kinetic energy formula:
\[ 4.806 \times 10^{-19} \text{ J} = \frac{p^2}{2 \times 9.11 \times 10^{-31}} \]
Solving for \(p\):
\[ p^2 = 2 \times 9.11 \times 10^{-31} \text{ kg} \times 4.806 \times 10^{-19} \text{ J} \]
\[ p^2 \approx 8.745 \times 10^{-49} \text{ kg}^2\text{m}^2/\text{s}^2 \]
\[ p \approx \sqrt{8.745 \times 10^{-49}} \text{ kg}\text{m}/\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3fe8677b-d2b4-4cf1-b1ed-08820154fcb5%2F6ee7da0c-b436-4de3-a58b-498db4225c7c%2Fpvlb8jq_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
An electron has a kinetic energy of 3.00 eV. Find its wavelength.
**Explanation:**
To find the wavelength of an electron with a given kinetic energy, one can use the de Broglie wavelength formula. The de Broglie wavelength (\(\lambda\)) for a particle is given by:
\[\lambda = \frac{h}{p}\]
where
- \(h\) is the Planck constant (\(6.626 \times 10^{-34} \text{ Js}\)),
- \(p\) is the momentum of the particle.
First, we need to find the momentum (\(p\)). The kinetic energy (KE) of an electron can be related to its momentum using the following equation:
\[ KE = \frac{p^2}{2m} \]
where \(m\) is the mass of the electron (\(9.11 \times 10^{-31} \text{ kg}\)).
Given \(KE = 3.00 \text{ eV}\), we need to convert this energy into Joules (since the Planck constant is in Joules·second). The conversion factor is:
\[ 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \]
So, \(3.00 \text{ eV}\) is:
\[ 3.00 \text{ eV} \times 1.602 \times 10^{-19} \text{ J/eV} = 4.806 \times 10^{-19} \text{ J} \]
Using the kinetic energy formula:
\[ 4.806 \times 10^{-19} \text{ J} = \frac{p^2}{2 \times 9.11 \times 10^{-31}} \]
Solving for \(p\):
\[ p^2 = 2 \times 9.11 \times 10^{-31} \text{ kg} \times 4.806 \times 10^{-19} \text{ J} \]
\[ p^2 \approx 8.745 \times 10^{-49} \text{ kg}^2\text{m}^2/\text{s}^2 \]
\[ p \approx \sqrt{8.745 \times 10^{-49}} \text{ kg}\text{m}/\
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 4 steps
