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De broglie wavelength of fast moving particle--help please!
A particle moving with kinetic energy equal to its rest energy has a de Broglie wavelength of 1.7898 *10^-16 m. If the kinetic energy doubles, what is the new de Broglie wavelength?
λ=h/√(2mK)
m=m0γ
K=Et-E0
Initially:
K=m0c2=ET-m0c2
But ET=m0c2γ
thus: 2=γ, which implies v=1/2(√3)
λ=h/√(2mK)=h/√2m0γ*m0c2=
h/(2m0c)
Then when the K becomes twice the rest mass:
K=Et-E0
3m0c2=m0c2γ, which implies γ=3
so λ=h/√(2m0γK)=h/√2m0*3*(2m0c2
=h/(2m0c*√3), which is just the original λ divided by the root of 3.
1.7898/√3= 1.03, so λ=1.033*10^-16, but the answer is λ=1.096*10^-16
Please help-- I have been wracking my brains about this for two days!
Homework Statement
A particle moving with kinetic energy equal to its rest energy has a de Broglie wavelength of 1.7898 *10^-16 m. If the kinetic energy doubles, what is the new de Broglie wavelength?
Homework Equations
λ=h/√(2mK)
m=m0γ
K=Et-E0
The Attempt at a Solution
Initially:
K=m0c2=ET-m0c2
But ET=m0c2γ
thus: 2=γ, which implies v=1/2(√3)
λ=h/√(2mK)=h/√2m0γ*m0c2=
h/(2m0c)
Then when the K becomes twice the rest mass:
K=Et-E0
3m0c2=m0c2γ, which implies γ=3
so λ=h/√(2m0γK)=h/√2m0*3*(2m0c2
=h/(2m0c*√3), which is just the original λ divided by the root of 3.
1.7898/√3= 1.03, so λ=1.033*10^-16, but the answer is λ=1.096*10^-16
Please help-- I have been wracking my brains about this for two days!