Electrons and the nucleus
I seriously didn't understand this question. If anyone can please explain it to me in detail. That will really be appreciated!
Original post by Daniel Atieh
I seriously didn't understand this question. If anyone can please explain it to me in detail. That will really be appreciated!
See post 27 in this thread:
http://www.thestudentroom.co.uk/showthread.php?t=104208&page=2
Using the help of ^ use lambda=h/p to get the electron's momentum with lambda=10^-15. This is because, as approximates, nucleus diameter = 10^-15 and atomic diameter = 10^-10. The question does seem to expect us to show a very high energy. So despite the very high speed in that post, it may be needed to calculate the electron's KE using KE=(p^2)/2m and maybe even give the answer in eV as you get an incredibly huge number if you work out KE in eV, try it.
Original post by krisshP
See post 27 in this thread:
http://www.thestudentroom.co.uk/showthread.php?t=104208&page=2
Using the help of ^ use lambda=h/p to get the electron's momentum with lambda=10^-15. This is because, as approximates, nucleus diameter = 10^-15 and atomic diameter = 10^-10. The question does seem to expect us to show a very high energy. So despite the very high speed in that post, it may be needed to calculate the electron's KE using KE=(p^2)/2m and maybe even give the answer in eV as you get an incredibly huge number if you work out KE in eV, try it.
http://www.thestudentroom.co.uk/showthread.php?t=104208&page=2
Using the help of ^ use lambda=h/p to get the electron's momentum with lambda=10^-15. This is because, as approximates, nucleus diameter = 10^-15 and atomic diameter = 10^-10. The question does seem to expect us to show a very high energy. So despite the very high speed in that post, it may be needed to calculate the electron's KE using KE=(p^2)/2m and maybe even give the answer in eV as you get an incredibly huge number if you work out KE in eV, try it.
Thanks a lot, man! ))
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