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Q.
What is de-Broglie wavelength of electron having energy 10 keV?
ManipalManipal 2008Dual Nature of Radiation and Matter
Solution:
de-Broglie wavelength of a particle is given by $ \lambda =\frac{h}{mv} $ ... (i)
where $ h $ is Plancks constant. If kinetic energy of particle of mass m is $ v $ , then $ K=\frac{1}{2}m{{v}^{2}} $
$ \Rightarrow $ $ v=\sqrt{\frac{2K}{m}} $ ... (ii)
Combining Eqs. (i) and (ii), we get
$ \lambda =\frac{h}{m\sqrt{\frac{2K}{m}}}=\frac{h}{\sqrt{2mK}} $ ... (iii)
Given: $ m=9.1\times {{10}^{-31}}\,\,kg $
$ K=10\,\,keV=10\times {{10}^{3}}\times 1.6\times {{10}^{-19}}J $
$ h=6.6\times {{10}^{-34}}\,\,J-s $
Substituting the above values in Eq. (iii), we get
$ \lambda =\frac{6.6\times {{10}^{-34}}}{\sqrt{2\times 9.1\times {{10}^{-31}}\times 10\times {{10}^{3}}\times 1.6\times {{10}^{-19}}}} $
$=1.22\times {{10}^{-11}} $
$ \approx 0.12\,\,\overset{\text{o}}{\mathop{\text{A}}}\, $
Note: If an electron is accelerated through a potential difference of $ V $ volt, then Eq. (iii) takes the form $ \lambda =\frac{h}{\sqrt{2meV}} $ After putting the numerical values for electrons, we get $ \lambda =\sqrt{\frac{150}{V}}\overset{\text{o}}{\mathop{\text{A}}}\, $