• # question_answer (i) A particle moving with velocity $5\times {{10}^{6}}m/s$ has de-Broglie wavelength of 0.135 nm associated with it. Calculate its kinetic energy in eV. (ii) In which region of the electromagnetic spectrum, does this wavelength lie?

(i) From de-Broglie matter wave equation, $\lambda =h/mv\Rightarrow m=h/\lambda v$ Here,     $\lambda =0.135\times {{10}^{-9}}m,$ $v=5\times {{10}^{6}}m/s$ $\therefore$ $m=\frac{6.63\times {{10}^{-34}}}{0.135\times {{10}^{-9}}\times 5\times {{10}^{6}}}=9.82\times {{10}^{-31}}kg$ $\therefore$ Kinetic energy of electron $K=\frac{1}{2}m{{v}^{2}}=\frac{1}{2}\times (9.82\times {{10}^{-31}}){{(5\times {{10}^{6}})}^{2}}$             $=4.91\times {{10}^{-31}}\times 25\times {{10}^{12}}$             $=122.75\times {{10}^{-19}}J$ $=\frac{122.75\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}ev=76.718\,eV$ (ii) This wavelength 0.135 nm falls in the region of X-ray of electromagnetic spectrum.