A) 33.28 nm
B) 3.328 nm
C) 0.3328 nm
D) 0.0332nm
Correct Answer: C
Solution :
Abiding to de-Broglie relation \[\lambda =\frac{h}{mv}=\frac{h}{p}\] \[\because \] \[p=\sqrt{2{{E}_{k}}.M}\] \[h=6.62\times {{10}^{-34}}kg{{m}^{2}}{{s}^{-1}}\] \[KE=13.6\times 1.602\times {{10}^{-19}}J\] \[KE=\frac{1}{2}m{{v}^{2}}=13.6\times 1.602\times {{10}^{-19}}J\] \[v=\sqrt{\frac{2\times 13.6\times 1.602\times {{10}^{-19}}}{9.1\times {{10}^{-31}}}}\] \[=2.18824\times {{10}^{6}}m/s\] \[\lambda =\frac{h}{mv}=\frac{6.626\times {{10}^{-34}}}{9.1\times {{10}^{-31}}\times 2.18824\times {{10}^{6}}}\] \[=0.3328\times {{10}^{-9}}\] or \[0.3328nm\]
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