What is the De Broglie Wavelength of 1.0-TeV Protons Accelerated at Fermilab?

[mramsey]

Homework Statement

De Broglie wavelength

What is the De Broglie wavelength of the 1.0-TeV (1TeV=10¹²eV)
protons (m_pc²=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

 

[creillyucla]

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