# De Broglie wavelength of 30keV electron

1. Dec 23, 2008

### catkin

1. The problem statement, all variables and given/known data
The question is from Advanced Physics by Adams and Allday, section 8 Practice Exam Questions, question 30.

Estimate the de Broglie wavelength of an electron that has been emitted thermionically in a vacuum from a filament and then accelerated through a p.d. of 30.0 kV

2. Relevant equations
λde Broglie = h / p
E2 - p2c2 = m02c4
ETotal = m0c2 + K.E.

3. The attempt at a solution
I think the solution is valid; my concern is whether there is a better (= more elegant) way to do it.

The de Broglie wavelength is given by
λde Broglie = h / p
Where h is Planck's constant and p is momentum.
p could be found from p = γm0v but this would require finding v. More conveniently
E2 - p2c2 = m02c4
Where E is the total energy (E = m0c2 + K.E.)
Expanding E:
p2c2 = 2m0c2K.E. + K.E.2
Rearranging:
p = (1/c) √(2m0c2K.E. + K.E.2)
Substituting this p:
λde Broglie = hc / √(2m0c2K.E. + K.E.2)
Substituting values using SI units (including eV to J conversion factor 1.60E-19)
= 6.63E-34 * 3.00E+8 / Sqrt(( 2 * 9.11E-31 * 3.00E+8^2 * 30E+3 * 1.60E-19) + (( 30E+3 * 1.60E-19)^2 ))
= 7.0e-12 m ct2sf

2. Dec 23, 2008

### Redbelly98

Staff Emeritus
Looks good.

By the way, this is a little easier using eV energy units. That way you won't have all those 1.6e-19's to contend with:

hc = 1240 eV-nm (a good number to remember in the future)
moc2 = 511 keV or 511e3 eV
KE = 30.0e3 eV

so that

λdB = 1240 eV-nm / √[2 × 511e3 × 30.0e3 eV2 + (30.0e3)2eV2]
= 0.00698 nm
= 6.98 pm ​

3. Dec 23, 2008

### Andrew Mason

Since it is asking only for an estimate, I would first determine whether the electron is moving at relativistic speeds. If it is not, you can use classical mechanics to determine p:

A 30 kV potential difference gives the electron 30keV of kinetic energy. Since it has about 500 keV as rest mass, the relativisitic effect will be small (since you are only estimating) so I would use a non-relativistic approach.

As suggested by Redbelly, use the formula:

$$\lambda_{DeB} = \frac{hc}{pc}$$

For non-relativistic speeds,

$$pc = \sqrt{2 KE m_0c^2}$$ (this is just a fancy rearrangment of $v = \sqrt{2 KE/m}$).

where KE is in units of eV. Note $m_0c^2$ = 511 KeV. and hc = 1240 eV nm

Use that to work out the Debroglie wavelength, in nm:

$$\lambda_{DeB} = hc/pc = 1240/\sqrt{2 x 3x10^4 x 5.11 x 10^5} = 1240/1.75 x 10^5 = 7 x 10^{-3}nm$$

AM

4. Dec 23, 2008

### Redbelly98

Staff Emeritus
Excellent point.

5. Jan 18, 2009

### catkin

Thanks Redbelly and Andrew.

Sorry it has taken me so long to get back here.

Helpful points, both about choice of unit and working out early on that the extra complexity of a relativistic approach is unnecessary.

What is the rest mass to K.E. ratio heuristic that allows deciding a relativistic approach is unnecessary and how does it arise?

Best

Charles

6. Jan 19, 2009

### Redbelly98

Staff Emeritus
Loosely speaking: when the KE is small compared to the rest mass energy, non-relativistic approximations may be used. Or equivalently, when v is small compared to c.

I'm not aware of any universally-accepted cutoff point, such as "when KE is x% of the rest mass energy". But as you saw, with KE equal to 6% of mc2, the non-relativistic result was pretty close, within 2%, of the actual de Broglie wavelength.

7. Feb 3, 2009

### catkin

Thanks Redbelly :)

The relativistic/Newtonian choice is something of a judgement call on this one. In our studies so far we have chosen 1% discrepancy in the result as the (arbitrary) determinant. On that basis, we found the cut of point for electrons is ~25keV acceleration. This is consonant with 2% error for an electron accelerated through 30 keV.

The thing that troubled me was that the question is from an old exam paper and the time to answer (based on the marks available) don't allow for my solution except the examinee be very good!

Perhaps that introduces an new relativistic/Newtonian choice determinant -- the time available to answer. :)

Best

Charles