solution:-
de-broglie wavelength can be written as
"\\lambda=\\frac{h}{p}=\\frac{h}{mv}=\\frac{h}{\\sqrt{2mKE}}"
therefore de-broglie wavelength for proton
"\\lambda _p=\\frac{h}{\\sqrt{2m_pKE_p}}...........eq.1"
de-brolie wavelength for electron can written as
"\\lambda _e=\\frac{h}{\\sqrt{2m_eKE_e}}...........eq.2"
from equation 1 and 2 we can write
"\\frac{\\lambda _p}{\\lambda _e}=\\sqrt{\\frac{m_e}{m_p}}"
because kinetic energy is same for proton and electron
"\\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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