De Broglie wavelength, kinetic energy

[ajmCane22]

Homework Statement

(a) Find the kinetic energy of an electron whose de Broglie wavelength is equal to 0.63 Å, a typical atomic size.

Ke = 379 eV

(b) Repeat part (a) for an electron with a wavelength equal to 1.7 x 10-15 m, a typical nuclear size.

Ke = _____eV

Homework Equations

Ke = h² / λ²(2m)

The Attempt at a Solution

I got part (a), but for part be I calculated Ke = 5.217e11 which is "significantly different" from the correct answer. I tried getting help from Cramster and they provided me with the same answer. What am I doing wrong? Please help!

 

[gneill]

Check the velocity that you calculate. How does it compare to c?

 

[ajmCane22]

I don't understand. I'm not calculating velocity? I'm calculating kinetic energy in Joules and converting it to eV. Where do c and v come in?

 

[gneill]

I don't understand. I'm not calculating velocity? I'm calculating kinetic energy in Joules and converting it to eV. Where do c and v come in?

When the energies get high enough the velocities will approach the speed of light. Then you need to take Relativistic effects into account. Check the velocity implied by the KE using Newtonian methods. If it's more than about 2/3 of c (or *gasp* greater than c!) you should be using the Relativistic formulas instead.

 

[asteriatic]

The difference in your answer for part (b) is most likely due to rounding errors. When dealing with very small numbers, it is important to use all significant figures in calculations to avoid such errors.

In part (b), the de Broglie wavelength is given in meters, so it is important to convert it to Ångstroms (Å) before plugging it into the equation. The correct answer for part (b) is Ke = 1.31 x 10-11 eV.

Also, when dealing with atomic and nuclear sizes, it is important to use the appropriate values for the mass of the electron (m) in the equation. For atomic sizes, the mass of the electron in the formula should be the rest mass (m0), while for nuclear sizes, it should be the relativistic mass (m0γ), where γ is the Lorentz factor. This is because the electron's velocity in a nuclear size is close to the speed of light, and its relativistic mass should be taken into account.

In summary, the correct way to solve part (b) would be:

1. Convert the given de Broglie wavelength from meters to Ångstroms. This can be done by multiplying by 10^10.

2. Use the appropriate mass of the electron for nuclear sizes, which is m0γ. The value for γ can be found using the formula γ = 1 / √(1 - (v/c)^2), where v is the velocity of the electron and c is the speed of light. In this case, since the electron's velocity is close to the speed of light, we can use the approximation γ ≈ 1.

3. Plug the values into the formula Ke = h2 / λ2(2m) and solve for Ke. You should get the correct answer of Ke = 1.31 x 10-11 eV.

I hope this helps! Remember to always pay attention to significant figures and use the appropriate values for mass in calculations involving atomic and nuclear sizes.