A) \[0.12\overset{\text{o}}{\mathop{\text{A}}}\,\]
B) \[1.2\overset{\text{o}}{\mathop{\text{A}}}\,\]
C) \[12.2\,\overset{\text{o}}{\mathop{\text{A}}}\,\]
D) None of these
Correct Answer: A
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 \[\lambda =\sqrt{\frac{150}{V}}\overset{\text{o}}{\mathop{\text{A}}}\,\]
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