Show change in de Broglie wavelength from change in speed

[Feynman.12]

Homework Statement

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

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.

 

[Orodruin]

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.

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

 

[Feynman.12]

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

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?

 

[BvU]

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

 

[Feynman.12]

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

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

 

[SammyS]

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

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

 

[Feynman.12]

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

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?

 

[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 $dy/y = -dx/x$ !

 

[Feynman.12]

$\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 $dy/y = -dx/x$ !

mindblow moment. Thankyou for your help!