[3dem] lambda

[Yifan Cheng]

Hi Natesh,

You need to take relativity into consideration. The "relativistic 
accelerating voltage" is V' = V0+(e/2mc*c)V0*V0.

Yifan

Ramanathan Natesh wrote:
> 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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>   

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Yifan Cheng, Ph.D.
University of California San Francisco
Department of Biochemistry and Biophysics
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