Relativistic De Broglie wavelength

  • Thread starter mramsey
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Homework Statement


De Broglie wavelength

What is the De Broglie wavelength of the 1.0-TeV (1TeV=1012eV)
protons (mpc2=938.3MeV) accelerated at the Fermilab Tevatron
accelerator? These high-energy protons are needed to probe
elementary particles [Hint: You need to use relativistic formula, but
consider potential simplification].


Homework Equations



λ = h / p
p=mv
λ = h /(mv*gamma)
hc = 1240 eV


The Attempt at a Solution



Using the relativistic form i then squared both sides to get λ2 = (h2 /(mv)2)(1-v2/c2)


From there i multiplied the two and multiplied by c2/c2 to get λ2 = (hc)2 /((mv2)(mc2)) - (hc)2 /((mc2)2)

I got stuck after that and not sure if i am doing it right i was thinking the mv2 could be the kinetic energy after multiplying by 1/2. λ2 = (hc)2 /2(((1/2)mv2)(mc2)) - (hc)2 /((mc2)2)and that the 1TeV is Kinetic Energy but that gives a negative number
 
Last edited:

Answers and Replies

  • #2
The easy equations to use here are

p = h / lambda

and E^2 = ( p * c )^2 + ( m c^2 )^2

you're pretty much done at that point
 

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