An alpha particle ( m = 6.64 × 10 − 27 m=6.64\times10^{-27} kg) emitted in the radioactive decay of uranium- 238 238 has an energy of 4.20 4.20 MeV. What is its de Broglie wavelength?

Hey, everyone. So this problem is working with wave particle duality. It is telling us that we have a release of neutrons. It's asking us to calculate the deli wavelength. So it's a big hint there um of that neutron. It's the problem is kind enough to give us the mass of the neutron and the kinetic energy of that neutron in this reaction. So the first thing that we need to remember is the debro wavelength debro wavelength equation. And that is um given as lambda equals H over P where P is the momentum, they don't give us momentum in this problem, but they do give us kinetic energy. So let's recall that relationship between kinetic energy and momentum where K equals P squared over to M where M is the mass. So we can solve this second equation in terms of P and then plug it into our first one. So that looks like two M times K, the square root of that equals P. So we can rewrite this to broadly wavelength equation as H over the square root of two MK. So let's double check that we have all of that um to plug in we recall the age planks constant is 6.63 times 10 to the negative 34 seconds. Um They give us the mass and the problem as 1.674 times 10 to the negative 27 kg. And then they do give us the kinetic energy, but it's given to us as four mega electron volts and planks constant is in jewels. So we're gonna have to do some unit work here to get this in the right format in the right unit. So mega electron volt is the same thing as saying, four times 10 GV six electron volt. And then we need to recall that one electron volt equals 1.602 times 10 to the minus 19 Jews. And so we can plug that in and that equals 6.41 times 10 to the minus 13 in Jules. So that is our kinetic energy of this neutron. So from there, we do have everything we need to solve for Delia wavelength. So we'll do that down here, links constant 6.63 times 10 to the minus 34 seconds square root, two times mass is 1.674 times 10 GB minus 27 kg. And then our kinetic energy in Jews is 6.41 times 10 to the minus 13 Jews. We plug that into our calculator and we get 1.43 times 10 to the minus 14 meters So let's look at our potential options and that aligns with the answer. A so that's all for this problem. We'll see you in the next video.

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Problem 4

Textbook Question

An alpha particle ( $m = 6.64 \times 10^{- 27}$ kg) emitted in the radioactive decay of uranium- $238$ has an energy of $4.20$ MeV. What is its de Broglie wavelength?

Verified step by step guidance

Convert the energy of the alpha particle from MeV to joules. Use the conversion factor: 1 MeV = 1.602 × 10^(-13) J. Multiply 4.20 MeV by this factor to get the energy in joules.

Determine the momentum of the alpha particle using the relationship between energy and momentum for a non-relativistic particle: $ E = \frac{p^2}{2m} $, where $ E $ is the energy, $ p $ is the momentum, and $ m $ is the mass. Rearrange the formula to solve for $ p $: $ p = \sqrt{2mE} $. Substitute the mass $ m = 6.64 \times 10^{-27} $ kg and the energy (in joules) into this equation.

Recall the de Broglie wavelength formula: $ \lambda = \frac{h}{p} $, where $ \lambda $ is the wavelength, $ h $ is Planck's constant ($ h = 6.626 \times 10^{-34} $ J·s), and $ p $ is the momentum. Use the momentum calculated in the previous step to find $ \lambda $.

Substitute the values of Planck's constant $ h $ and the momentum $ p $ into the de Broglie wavelength formula to calculate $ \lambda $. Ensure the units are consistent throughout the calculation.

Express the final de Broglie wavelength in meters (m) and, if needed, convert it to a more convenient unit such as nanometers (nm) by using the conversion factor: 1 m = 10^9 nm.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

De Broglie Wavelength

The de Broglie wavelength is a fundamental concept in quantum mechanics that relates a particle's momentum to its wavelength. It is given by the formula λ = h/p, where λ is the wavelength, h is Planck's constant (6.626 x 10^-34 Js), and p is the momentum of the particle. This concept illustrates the wave-particle duality of matter, indicating that particles like alpha particles exhibit both particle-like and wave-like properties.

Energy and Momentum Relationship

In physics, the energy of a particle is related to its momentum through the equation E = p^2/2m for non-relativistic particles, where E is energy, p is momentum, and m is mass. For an alpha particle, which is a type of helium nucleus, its kinetic energy can be converted into momentum, allowing us to calculate its de Broglie wavelength. Understanding this relationship is crucial for solving problems involving particle behavior in quantum mechanics.

Conversion of Energy Units

In the context of this problem, it is important to convert energy units from MeV (mega-electronvolts) to joules for calculations involving the de Broglie wavelength. The conversion factor is 1 MeV = 1.602 x 10^-13 joules. This conversion is necessary because the standard units in physics equations typically use joules, ensuring consistency and accuracy in calculations involving energy, momentum, and wavelength.

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