[3dem] lambda
Radostin Danev
rado at nips.ac.jp
Thu Nov 15 17:33:08 PST 2007
Hi Natesh,
Here is a simple way to derive lambda:
lambda = h/p # see here:
http://en.wikipedia.org/wiki/De_Broglie_hypothesis)
# h is the Planck constant, p is the relativistic
momentum
p*c = sqrt(E^2-m0^2*c^4) # see here:
http://hyperphysics.phy-astr.gsu.edu/hbase/relativ/relmom.html)
# m0 is the rest mass of the electron, E is the
total energy
E = m0*c^2 + e*U # e is the unit charge, U is the acceleration
voltage
substitute the equations upwards and you get:
lambda = h*c/sqrt(2*m0*c^2*e*U + e^2*U^2)
substitute all constants and you'll get the formula you needed.
Best regards,
Rado
-----Original Message-----
From: 3dem-bounces at ncmir.ucsd.edu [mailto:3dem-bounces at ncmir.ucsd.edu] On
Behalf Of Ramanathan Natesh
Sent: Friday, November 16, 2007 03:18
To: 3dem at ucsd.edu
Subject: [3dem] lambda
Hi All,
Could anyone point me to a reference or show me how do we get from
classical formula for wavelength
lambda = h/sqrt(2meV) to lambda = 12.398/sqrt[kV*(1022+kV)]
that is used in SPIDER glossary
http://www.wadsworth.org/spider_doc/spider/docs/glossary.html#Lambda
m=mass 9.11 x 10^-31 Kg; e=electron charge 1.6 x 10^-19 C
h= 6.62 x 10^-34 J*sec V=voltage in volts
kV=kilo Volts.
eg: for 120 kV I get lambda = 0.03539 using classic formula but
I get lambda = 0.03349 from formula in SPIDER
glossary. Though the variation in 3rd decimal may/may not be
significant.
thanks in advance,
natesh.
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