Answer to Question #132350 in Mechanics | Relativity for Viccy

Question #132350
What is ratio of De-broglie wavelegth of proton and electron at same kinetic energy.
1
Expert's answer
2020-09-10T11:59:01-0400

De Broglie wavelength is

λ=hp\lambda = \frac{h}{p}

E=p22mp=2mEE = \frac{p^2}{2m} \Rightarrow p = \sqrt{2mE}

λpλe=pepp=2meEe2mpEp\frac{\lambda_p}{\lambda_e} = \frac{p_e}{p_p} = \frac{\sqrt{2m_eE_e}}{\sqrt{2m_pE_p}}

Ee=EpE_e = E_p (same kinetic energy), so

λpλe=memp=9.110311.671027=5.449104=2.334102=0.023\frac{\lambda_p}{\lambda_e} = \frac{\sqrt{m_e}}{\sqrt{m_p}} = \sqrt{\frac{9.1 \cdot 10^{-31}}{1.67 \cdot 10^{-27}}} = \sqrt{5.449 \cdot 10^{-4}} = 2.334 \cdot 10^{-2} = 0.023


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