De Broglie wavelength is
λ=hp\lambda = \frac{h}{p}λ=ph
E=p22m⇒p=2mEE = \frac{p^2}{2m} \Rightarrow p = \sqrt{2mE}E=2mp2⇒p=2mE
λpλe=pepp=2meEe2mpEp\frac{\lambda_p}{\lambda_e} = \frac{p_e}{p_p} = \frac{\sqrt{2m_eE_e}}{\sqrt{2m_pE_p}}λeλp=pppe=2mpEp2meEe
Ee=EpE_e = E_pEe=Ep (same kinetic energy), so
λpλe=memp=9.1⋅10−311.67⋅10−27=5.449⋅10−4=2.334⋅10−2=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} = 0.023λeλp=mpme=1.67⋅10−279.1⋅10−31=5.449⋅10−4=2.334⋅10−2=0.023
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