A) \[3.3\times {{10}^{-6}}m\]
B) \[3.3\times {{10}^{5}}m\]
C) \[3.3\times {{10}^{-4}}m\]
D) 30 m
Correct Answer: A
Solution :
From de-Broglie's relation, we have \[\text{wavelength ( }\!\!\lambda\!\!\text{ ) = }\frac{\text{planck }\!\!'\!\!\text{ s}\,\text{cosntant (h)}}{\text{momentum (p)}}\] Putting the numerical values from the question , we have \[\lambda =\frac{6.6\times {{10}^{-34}}}{2\times {{10}^{-28}}}=3.3\times {{10}^{-6}}\,\text{m}\]
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