Given the wavelength (250nm) of an electron, find its\' energy. I know that the

Question

Given the wavelength (250nm) of an electron, find its' energy.

I know that the correct derivation of the equation is this...

E = K = 1/2mv^2 = (mv)v/2 = [(mv)*(mv)]/2m = (mv)^2/2m = p^2/2m

Since = h/p, the p = h/

E = P^2/2m = (h/)^2/2m

E = h^2/2m^2 <-- Correct equation

But I got the following before I looked up the answer...

E = K = 1/2mv^2

v^2 =2E/m

E = hf; de Broglie's equation

Since f = v/. then

E^2 = 2Eh^2/m^2

E = 2h^2/m^2 <--- Wrong equation

I don't see what I did wrong. Iknow that squaring both sides can introduce extreaneous roots,but I am off by a constant factor of 4.

I have broken some rule of algebra, but I do not see it.

Explanation / Answer

in E=hf u replaced f with v/lamda but it should be c/lamda http://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality check about the wavelength formula once

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