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if the total energy of an electron in a hydrogen like atom in excited state is -3.4 eV then the de-broglie wavelength of the electron is
total energy in ground state (t.e) = -13.6ev hence, n^2 = 13.6/3.4 n=2 v(velocity)= 2pkze^2/nh hence, v= 2×3.14×9×10^9×1×(1.6× 10^-19)^2 ÷ 2×6.626 ×10^-34 hence,v=10.96×10^5 p(linear momentum)= mv de broglie wavelength(wl) =h/p wl= 6.626×10^-34/9.1×10^-31×10.96×10^5 wl=0.06643× 10^-8 wl= 6.643angstorm
Total Energy K = 3.4 eV = 3.4x1.9x10^-19 joule.So p = Sqrt (2mK) = sqrt (2*9.1*10^-31*3.4x1.9x10^-19) = 10/84x10^-35 kgm/sSo debroglie wavelength = h/p = 6.63x10^-34 / 10.84x10^-25 = 0.611x10^-9 = 0.611 nmRegards,Nirmal SinghAskiitians faculty
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