(a) An electron has a kinetic energy of 3.21 eV. Find its wavelength. nm (b) A p

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Explanation / Answer

(A)Energy=hc/Lamda c=3*(10^8)m/s h=(6.625) * (10^-34) J-se Lamda=wavelenght=(hc)/Energy ==> wavelenght={(3*10^8)(6.625*10^-34)}/{3.21*1.6*(10^-19)} ==>wavelenght=3.86*(10^-7)m=386.9nm (B)E = (1/2) mv^2 = (mv)^2 / (2m = p^2/2m => p = v(2mE) de Broglie wavelength, lamda= h/p Mass of proton = 1.67262158 × 10-27 kilograms Planck's constant, h = 6.626068 × 10-34 m^2 kg / s K.E. given = 3.21 eV = 3.21 x 1.60217646 × 10-19 joule=5.136*(10^-19) ==>p=4.14*(10^-23) lamda=1.599*(10^-11)m=0.0159nm

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