Solve de Broglie Wavelength of Atoms in eV: Urgent Help

In summary, the equation for the homework statement is: lambda = h/p. Homework Statement The equation for the homework statement is: lambda = h/p.
  • #1
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Homework Statement


http://www.strings.ph.qmul.ac.uk/~russo/QP/r06QPHYEX.pdf [Broken]
Question 3.

The Attempt at a Solution


I can easily derive the equation I need, which is:

lamda = h/p

Which after some playing around with K.E =1/2mv^2 etc we obtain:

lambda (de brogile) = h/SQRT(2mKE)

Here is the data we are provided with in the exam (see sheet 1): http://www.strings.ph.qmul.ac.uk/~russo/QP/week6.pdf [Broken]

Everythings in eV and MeV, I plug in the values constantly, tried converting h to joules, and the electron mass to kg etc, but no avail as I don't think I know what I am doing with it, and I can't seem to obtain the real answer which is: 1.2 x10^-9m

Anyone explain how to do this quickly? I understand the theory behind this top notch but missed on how to use eV etc.

THANKS.
 
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  • #2
The equation appears to be right (I plugged in values and got the right answer), what kind of answers are you getting? It's probably just an arithmetic error, or a unit error.
 
  • #3
Thanks for the reply!

I plugged in:
4.14x10^-21 / SQRT(2 x 0.511 x 1.6x10-24) = 3.2 x 10-9.

I converted 1eV into MeV. Where am I going wrong with this?
If I convert everything to eV and use:
4.14x10^-15 / SQRT(2 x 511 x 1.6x10-19) = 3.2 x 10-7m

Still wrong, and I'm ripping my hair out on this!
 
  • #4
Well, when you use electron volts, what speed comes out? It's not meters/second, but rather c.

Also, you have some wrong values. The electron is 511keV = .511MeV (correct for your calculation in MeV) = 511000eV (incorrect when you plugged it in eV).

Also, you're converting 1 eV into joules which doesn't make sense since you used everything else in eV.
 
  • #5
So how would I go about doing this?

Would I multiply the value of the rest mass by (3x10^8)^2 to obtain the MeV value?
 
  • #6
no, because then you get a dimensionless unit (or /s or something weird like that).

What you want to realize is first, don't convert your 1eV into 1.602*10^-19 Joules, and 2 at the end realize that your answer is given as a fraction of c (times seconds).

How would you then convert the answer to the correct one? Well, multiply your answer by the value of c.

The units will be weird, but the procedure should get you the right answer at the end.
 

1. What is the de Broglie wavelength of an atom in eV?

The de Broglie wavelength is a concept in quantum mechanics that describes the wavelength of matter, including atoms, as a wave. It is calculated using the equation λ = h/mv, where h is Planck's constant, m is the mass of the particle, and v is its velocity. When the wavelength is measured in units of electron volts (eV), it is known as the de Broglie wavelength of an atom in eV.

2. How is the de Broglie wavelength of an atom in eV different from its wavelength in meters?

The de Broglie wavelength of an atom in eV is a unit of measurement commonly used in quantum mechanics, while its wavelength in meters is a more traditional unit of length. The two measurements are related by the conversion factor 1 eV = 1.97x10^-7 meters. This means that the de Broglie wavelength of an atom in eV will be much smaller than its wavelength in meters.

3. Can the de Broglie wavelength of an atom in eV be measured experimentally?

Yes, the de Broglie wavelength of an atom in eV can be measured experimentally using diffraction techniques. When atoms are fired at a diffraction grating, their wave-like behavior can be observed, and the de Broglie wavelength can be calculated from the diffraction pattern. This has been successfully demonstrated with electrons, neutrons, and even large molecules like buckyballs.

4. How does the de Broglie wavelength of an atom in eV relate to its energy?

The de Broglie wavelength of an atom in eV is inversely proportional to its energy. This means that as the energy of the atom increases, its de Broglie wavelength decreases. This relationship is consistent with the wave-particle duality of matter, where high-energy particles behave more like particles, while low-energy particles behave more like waves.

5. Why is it important to calculate the de Broglie wavelength of an atom in eV?

Calculating the de Broglie wavelength of an atom in eV is important because it helps us understand the wave-like behavior of particles in the quantum world. It also allows us to make predictions about the behavior of matter at the atomic scale, which is crucial for many technological applications, such as nanotechnology and quantum computing.

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