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?

Answer:

(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.