A) \[{{10}^{3}}h\] metre
B) \[{{10}^{5}}h\]metre
C) \[{{10}^{-3}}h\]metre
D) \[{{10}^{-5}}h\]metre
Correct Answer: D
Solution :
: de Broglie wavelength\[(\lambda )=\frac{h}{mv}\] or\[\lambda =\frac{h}{({{10}^{-6}})\times 1}\]since \[1\text{ }mg={{10}^{-6}}kg\] or \[\lambda ={{10}^{6}}h\] metre.
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