Asked by Kyle
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]
What is the de Broglie wavelength of this electron?
[KE= (1/2)mv^2; 1 electron volt (eV) = 1.602*10^-19 J]
Answers
Answered by
bobpursley
p= sqrt (KE*m*2)
= sqrt(505*1.602E-19*9.11E-31*2)
then
lambda= plancks constant/p
= sqrt(505*1.602E-19*9.11E-31*2)
then
lambda= plancks constant/p
Answered by
Anonymous
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?
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.