A) 0.032 nm
B) 0.40 nm
C) 2.5 nm
D) 14.0 nm
Correct Answer: B
Solution :
\[\lambda =\frac{h}{mv}=\frac{6.63\times {{10}^{-34}}}{1.67\times {{10}^{-27}}\times 1\times {{10}^{3}}}\] \[=\frac{6.63\times {{10}^{-34}}}{1.67\times {{10}^{-24}}}\] \[=0.40\times {{10}^{-9}}m\] \[=0.40\text{ }nm\]
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