A) 95pm
B) 102pm
C) 112pm
D) 124pm
Correct Answer: C
Solution :
The de-Broglie wavelength of a electron of \[{{m}_{e}}\]and kinetic energy K is given by \[\lambda =\frac{h}{\sqrt{2\,{{m}_{e}}\,K}}\] Here, \[{{m}_{e}}=9.1\times {{10}^{-31}}kg\] and \[K=120\,eV=120\times 1.6\times {{10}^{-19}}J\] \[\therefore \] \[\lambda =\frac{6.6\times {{10}^{-34}}}{\sqrt{2\times 9.1\times {{10}^{-31}}\times 120\times 1.6\times {{10}^{-19}}}}\] \[=\frac{6.6\times {{10}^{-34}}}{\sqrt{3494.4\times {{10}^{-50}}}}\] \[\lambda =\frac{6.6\times {{10}^{-34}}}{59.1\times {{10}^{-25}}}\] \[\lambda =112.0\times {{10}^{-12}}m\] \[\lambda =112\,pm\]
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