## test 2, question5

$\lambda=\frac{h}{p}$

Rosamari Orduna 1D
Posts: 31
Joined: Fri Apr 06, 2018 11:04 am

### test 2, question5

During a reflex signal, the local velocity of a sodium ion is assumed to be 112m/s. What is the de Broglie wavelength of one ion of sodium?

So ive set up my equation: (6.626x10^-34J.s) / (22.989g/mol^-1)(112m/s)(6.022x10^23mol^-1)(1000g/1kg)

multiplied out to: (6.626x10^-34J.s) / (1.55x10^30m.s.kg)= 4.27x10^-64m

to me it just seems like the exponent is wayyyy too large, but thats just me doubting myself, but i could be wrong, or rightt...?

also, this question has been answered before, for a different version, but i just need to check my answer.

Rosamari Orduna 1D
Posts: 31
Joined: Fri Apr 06, 2018 11:04 am

### Re: test 2, question5

hahaha sorry to all, I think i figured out my error because that number seemed wayyy too small. I was multiplying everything out together on my calculator and I think that caused it to give me the wrong numbers and had some cancellation errors.

If someone could please confirm if my answer is correct: 1.74X10^-11m is the wavelength size. I got this by (6.626x10^-34) / (3.8x10^-23)

i hope this is righttt...

Jesus A Cuevas - 1E
Posts: 29
Joined: Fri Apr 06, 2018 11:02 am

### Re: test 2, question5

Hey, so I am still having a little trouble with this problem. I know we must use the equation λ=(h/p) = (h/ (m*v)), but could anyone clarify how to solve this problem? Thanks in advance!

Posts: 41
Joined: Wed Nov 15, 2017 3:01 am

### Re: test 2, question5

Why are suppose to use Avogadro's number for this problem?

Kuldeep Gill 1H
Posts: 44
Joined: Fri Apr 06, 2018 11:02 am
Been upvoted: 1 time

### Re: test 2, question5

You use Avogadro's number to find the mass of a sodium ion since sodium ions have mass and in order to find de Broglie wavelength, we need both mass and velocity.

tmehrazar
Posts: 39
Joined: Fri Apr 06, 2018 11:04 am

### Re: test 2, question5

I'm not sure, but I got 1.550 x 10^-10m and I got that from (6.626x10^-34Js)/(3.818x10^-26kg)(112ms^-1)

I got 3.818x10^-26kg by multiplying 22.99g Na/1mol Na x 1mol Na/6.022x10^23 x 1kg/1000g

Not 100% I'm right though

Can anyone validate this?