p= sqrt (KE*m*2)
= sqrt(505*1.602E-19*9.11E-31*2)
then
lambda= plancks constant/p
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]
2 answers
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?
1 answer