Question

Those who like Quantum Mechanics :
Join the post :)

Need help :)

Answer 1

An Electron has a of speed 400m/s with uncertainty of 0.01% . Find the accuracy in its position.

Answer 2

@IrishBoy123 @Michele_Laino @ganeshie8 @Kainui

Answer 3

Are you familiar with the Heisenberg uncertainty relation?
\[\Delta x \Delta p \ge \frac{\hbar}{2}\]

Answer 4

h* ig

Answer 5

is it divided by 2 ?

Answer 6

I've often seen people drop the 1/2, it's not wrong because the Heisenberg uncertainty relation is already an approximation anyways, so since this is true:

\[\Delta x \Delta p \ge \hbar \ge \frac{\hbar}{2}\]

It's not wrong to say this either. I mean it's whatever your book/teacher/class wants really.

Answer 7

hahah true
whatever the teacher wants :)

Answer 8

I can show you the derivation but I imagine it's probably not going to be very enlightening to you at this point since it's buried in Fourier transforms lol

Answer 9

no
lol
not now

Answer 10

\(\Delta x\) is really the root mean square deviations from the average. So if I use \(\langle a \rangle\) to mean "the average value of a" then:

\[\Delta x = \sqrt{\langle (\langle x \rangle - x)^2 \rangle }\]

I think this is also called the "standard deviation"... But anyways you want to get some answers, throw up some equations you think might be useful from your class's notes or your book or whatever so we can do this consistent with what they expect.

Answer 11

lol
i got the solution i guess

\(\rm p=mv\\~~~~~~~ 9.11~X~10^{-31}~X~ 400\\~\\~ \Delta p=m \Delta V=m v\frac{\Delta v}{v} \\\)

Answer 12

btw
could u explain the basic math of Delta V --> deltav/v

Answer 13

Yeah, definitely this is the trick you need to use the information you were given.

You see, they give you the velocity v, and then they give you the percent uncertainty which is 100 off from this quantity here: \(\frac{\Delta v}{v}\)

When they introduce it into the formula, they're using the fact that \(\frac{v}{v} = 1\) so you can sneak it in.

Answer 14

OOHH

Answer 15

so then
after that the HUP :)

Answer 16

Yeah you got it, make sure you use what they use, whether it's \(\Delta x \Delta p \ge h\) or \(\Delta x \Delta p \ge \tfrac{\hbar}{2}\) or whatever haha

Answer 17

well wait
uncertainty is \(\Delta v/v \)

Answer 18

and \(\Delta p = m\Delta V\)

Answer 19

so \(\Delta V =?\)

Answer 20

No, uncertainty is \(\Delta v\) and percent uncertainty is \(\frac{\Delta v}{v}*100\)

Answer 21

so i could just put in m \(\Delta\)v /v * 100

Answer 22

anyways i got it :)
YAY!

Answer 23

lol alright good

Answer 24

Thank you :)

Answer 25

kai

Answer 26

i got confused
m \(\Delta V\)

Answer 27

An Electron has a of speed 400m/s with uncertainty of 0.01% . Find the accuracy in its position.

In symbols, what are each of these numbers referring to?

Answer 28

v=400
\(\Delta\) v=0.01
\(\Delta\)x =?

Answer 29

% uncertainty = \(\Large\rm\frac{\Delta v}{100}\cdot v\)

Answer 30

Yeah what you wrote: \(\Delta v = 0.01\) is wrong, but this second post is looking lik eyou're heading in the right direction:

\[\text{uncertainty of 0.01% }= \frac{\Delta v}{v} * 100\]
Or in other ways of writing:
\[.0001 = \frac{\Delta v}{v}\]

Answer 31

yus!
i got it now !!!!!!!!!!!!!!!
Yay :)