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Match Column - I with Column - II : |
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Answer» `sqrt(2mE) = mv` de Broglie wavelength , `lamda=h/(mv)=h/(sqrt(2mE))` Momentum of a particle and its kinetic energy (E) are related as :de Broglie wavelength , When kinetic energy of alpha particle is found TIMES GREATER than that of alpha particle and momentum of alpha particle is greater than that of proton. When kinetic energy of alpha particle is one fourth pf proton , `lamda_alpha/lamda_p=sqrt((m_pE_p)/(m_alphaE_alpha))=1` Thus wavelength for both the particles is same. However in case of momentum , `lamda_alpha/lamda_p=sqrt((m_p)/(m_alpha))=sqrt(1/4)=1/2` Thus, momentum of proton is less than that of alpha particle . When velocity of alpha particles is equal to that of proton , their energy ratio will be `E_(alpha)/(E_p)=((m_alphav_alpha^2)/2)/(m_pv_p^2)=m_alpha/m_p=4/1` `:.E_alpha=4E_p` Thus wavelength of photon isgreater than that of alph particle and momentum of alpha particle is greater than that of photon. Similarly, when kinetic energy of alpha particle is FOUR times greater than that of protons , than wavelength of proton is greater than of alpha particle and momentum of alpha particle is greater than that of proton. |
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