[3dem] lambda

[Radostin Danev]

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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