# Show change in de Broglie wavelength from change in speed

Tags:
1. Jul 14, 2015

### Feynman.12

1. The problem statement, all variables and given/known data

Show that for a nonrelativistic particle, a small change in speed leads to a change in de Broglie wavelength given from

3. The attempt at a solution

I have tried to expand the left hand side of the equation, but found that it gave the answer of v0/delta v. My definition of delta lambda is the final wavelength minus the initial wavelength.

2. Jul 14, 2015

### Orodruin

Staff Emeritus
You need to actually show us what you did. How else can we find out where and if you went wrong?

3. Jul 14, 2015

### Feynman.12

Sorry, my attempt is as follows.

In the book (Eisberg, Resnick - quantum physics of atoms, molecules, solids, nuclei and particles, pg. 82, question 10) it has the answer as that given above, however, my attachment proves that wrong. Is there anywhere I may have made a mistake?

#### Attached Files:

File size:
49.2 KB
Views:
215
• ###### image.jpg
File size:
49.2 KB
Views:
185
4. Jul 14, 2015

### BvU

In your first step you write something that looks like ${1\over a -b} = {1\over a} - {1\over b}$ to me ...

5. Jul 14, 2015

### Feynman.12

I can't find where I have done this. How would you do this question?

6. Jul 15, 2015

### SammyS

Staff Emeritus
What you did is equivalent to that.

You had $\displaystyle \ \Delta\lambda=\frac{h}{mv_f}-\frac{h}{mv_i} \ .$

Then you did this:
$\displaystyle \ \frac1{\Delta\lambda}=\frac{mv_f}{h}-\frac{mv_i}{h} \ .$

However, $\displaystyle \ \frac1{\displaystyle\frac{h}{mv_f}-\frac{h}{mv_i}}\ne\frac{mv_f}{h}-\frac{mv_i}{h} \ .$

#### Attached Files:

File size:
66.6 KB
Views:
119
7. Jul 15, 2015

### Feynman.12

Okay, I understand that! I was able to try and attempt to solve this with the new knowledge, however I got stuck. I derive an answer that is

$\displaystyle \ \frac{\Delta \lambda}{\lambda_0}=\frac{-\Delta v}{v_f}$
-

If my mathematics is correct, that would mean that

$\displaystyle \ \frac{-\Delta v}{v_f}=\frac{\Delta v}{v_0}$

But I can't think of any relations that would make the above true?

8. Jul 15, 2015

### BvU

$\displaystyle \ \frac{\Delta \lambda}{\lambda_0}=\frac{-\Delta v}{v_f}\ \$ is correct. The book means to say $\displaystyle \ \frac{\Delta \lambda}{\lambda}=\frac{|\Delta v|}{v}\ \$ but finds the sign so trivial that it leaves out the $|\ |$.
And for a small change $v = v_0 \approx v_f$ in the denominator -- NOT, of course in the difference.

This reminds me of differentiation and error propagation:

With $y = 1/x$ you have $dy = -1/x^2 \; dx$ so $dy/y = -dx/x$ !

9. Jul 15, 2015

### Feynman.12

mindblow moment. Thankyou for your help!