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Photon having energy equivalent to the binding energy of 4th state of He^(+) ion is used to eject an electron from the metal with KE/2eV . If electron is further accelerated through a potential difference of 4V then the minimum value of de Broglie wavelength associated with the electron is : (h = 6.6 xx 10^(-34) J-s , m_(e) = 9.1 xx 10^(-31) kg . 1 eV = 1.6 xx 10^(-19) J )

Refer to illustration 5.10 , the energy released by the alpha - decay of ._(92)^(238) U is found to be 4.3 MeV . Since this energy is carried away as kinetic energy of the recoiling ._(90)^(234)Th . nucleus and the alpha - particles, it follows that KE_(Th) +KE_(alpha )=4.3 MeV . However, KE_Th and KE_alpha are not equal. Which particle carries away more kinetic energy, the overset(234)(90)Th nucleus or the alpha - particle? .