Posted by **Kyle** on Friday, November 13, 2009 at 3:37pm.

The presence of oxygen atoms on the surface results in auger electrons with a kinetic energy of approximately 505 eV.

What is the de Broglie wavelength of this electron?

[KE= (1/2)mv^2; 1 electron volt (eV) = 1.602*10^-19 J]

- de Broglie Wavelength -
**bobpursley**, Friday, November 13, 2009 at 4:21pm
p= sqrt (KE*m*2)

= sqrt(505*1.602E-19*9.11E-31*2)

then

lambda= plancks constant/p

- de Broglie Wavelength -
**Anonymous**, Tuesday, July 5, 2011 at 1:34pm
What is the wavelength (in meters) of a proton ({\rm{mass}} = 1.673 \times 10^{ - 24} {\rm g}) that has been accelerated to 28% of the speed of light?

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