solution:-
de-broglie wavelength can be written as
λ=hp=hmv=h2mKE\lambda=\frac{h}{p}=\frac{h}{mv}=\frac{h}{\sqrt{2mKE}}λ=ph=mvh=2mKEh
therefore de-broglie wavelength for proton
λp=h2mpKEp...........eq.1\lambda _p=\frac{h}{\sqrt{2m_pKE_p}}...........eq.1λp=2mpKEph...........eq.1
de-brolie wavelength for electron can written as
λe=h2meKEe...........eq.2\lambda _e=\frac{h}{\sqrt{2m_eKE_e}}...........eq.2λe=2meKEeh...........eq.2
from equation 1 and 2 we can write
λpλe=memp\frac{\lambda _p}{\lambda _e}=\sqrt{\frac{m_e}{m_p}}λeλp=mpme
because kinetic energy is same for proton and electron
λpλe=9.1×10−311.67×10−27=0.021\frac{\lambda _p}{\lambda _e}=\sqrt{\frac{9.1\times10^{-31}}{1.67\times10^{-27}}}=\frac{0.02}{1}λeλp=1.67×10−279.1×10−31=10.02
therefore ratio of de-broglie waelength of proton and electron is 0.02:1
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