Help with light question - electron wavelength and planck's constant

[s2lindah]

Hi,

I'm having trouble with this question. "Calculate the deBroglie's wavelength for an electron with 66eV."
I used the formulas p = (2mE)^0.5 and wavelength = h/p, which gives wavelength = h/[(2mE)^0.5]

However, if you use that formula with Planck's constant in Js (6.63*10^-34), and also convert 66eV to J, you get a different answer than using Planck's constant in eVs (4.14*10^-15) and using 66eV.

Mass of electron is constant at 9.1*10^-31..

Why is it different? And which answer would be correct - which Planck's constant must we use and why?

Thanks!

[Whatlol]

The answers should not differ too much, but the reason they are different is because they are not entirely accurate.  for example if you convert 6.63x10^-34 into eVs you get 4.14375x10^-15 , yet usually that is just given to us as 4.14x10^-15. thats why the answers are different.


And whether to use one or the other im not actually sure. I always just stick with 6.63x10^-34.

[s2lindah]

Thanks for your fast reply

Nah the answers are almost completely different - with the Planck's constant as eVs and energy in eV, wavelength is 0.378m, whilst using Planck's constant as Js and energy in J, wavelength is 1.5*10^-10

[itolduso]

The E in p = (2mE)^0.5 must be in J.
So h must be in Js.

[s2lindah]

why must it be in J?

[/0]

You have to be very careful when doing calculations in "eV", and in VCE I recommend that you convert all "eV"s to "J"s before doing calculations (unless it's simple stuff). The reason why is if you use electronvolts, your units of mass change from to .

Consider:

The LHS has units of , so the RHS must also have units of .

If is in , is in , and is in , then we have:



Recall that since ,

So , as we would expect.


However, if we have in and in , then must have units of (electronvolts divided by the speed of light squared), since:




To convert from a mass in to a mass in , you multiply by then divide by

The mass of electron in is: .

Then, if u perform the calculation in electronvolts,

[Whatlol]

why must it be in J?

because Ek = 1/2mv^2 , the unit is not in ev , it is in joules.

to use 4.14x10^-15 you would have to do the following.
squre root of (2mEK / 1.6x10^-19 ) = p
therefore wavelength = root (1.6x10^-19) x 4.14x10^-15 / root (2mEK)

[ghadz7]

If you wanna use eV, time the value by root(1.6 x 10^-19)

You got 0.378 while using 66eV and 4.14 x 10^-15 however you ignored the fact that we are rooting the energy

0.378 x root(1.6 x 10^-19) and you get teh accurate answer of 1.51 x 10^-10

I suggest you always use Joules with these questions.

[Whatlol]

lol /0 nice job

[s2lindah]

Thankyou everyone for your help!!! Especially /0. That really clarified things