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## Homework Statement

De Broglie wavelength

What is the De Broglie wavelength of the 1.0-TeV (1TeV=10

^{12}eV)

protons (m

_{p}c

^{2}=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 λ

^{2}= (h

^{2}/(mv)

^{2})(1-v

^{2}/c

^{2})

From there i multiplied the two and multiplied by c

^{2}/c

^{2}to get λ

^{2}= (hc)

^{2}/((mv

^{2})(mc

^{2})) - (hc)

^{2}/((mc

^{2})

^{2})

I got stuck after that and not sure if i am doing it right i was thinking the mv

^{2}could be the kinetic energy after multiplying by 1/2. λ

^{2}= (hc)

^{2}/2(((1/2)mv

^{2})(mc

^{2})) - (hc)

^{2}/((mc

^{2})

^{2})and that the 1TeV is Kinetic Energy but that gives a negative number

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