Answer to Question #132235 in Mechanics | Relativity for Max

Question #132235
What is ratio of De-broglie wavelegth of proton and electron at same kinetic energy.
1
Expert's answer
2020-09-11T09:15:44-0400

solution:-

de-broglie wavelength can be written as

λ=hp=hmv=h2mKE\lambda=\frac{h}{p}=\frac{h}{mv}=\frac{h}{\sqrt{2mKE}}


therefore de-broglie wavelength for proton


λp=h2mpKEp...........eq.1\lambda _p=\frac{h}{\sqrt{2m_pKE_p}}...........eq.1


de-brolie wavelength for electron can written as


λe=h2meKEe...........eq.2\lambda _e=\frac{h}{\sqrt{2m_eKE_e}}...........eq.2



from equation 1 and 2 we can write


λpλe=memp\frac{\lambda _p}{\lambda _e}=\sqrt{\frac{m_e}{m_p}}


because kinetic energy is same for proton and electron


λpλe=9.1×10311.67×1027=0.021\frac{\lambda _p}{\lambda _e}=\sqrt{\frac{9.1\times10^{-31}}{1.67\times10^{-27}}}=\frac{0.02}{1}


therefore ratio of de-broglie waelength of proton and electron is 0.02:1




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