A) \[33.28\text{ }nm\]
B) \[3.328\text{ }nm\]
C) \[0.3328\text{ }nm\]
D) \[0.0332\text{ }nm\]
Correct Answer: C
Solution :
For electron in the ground state, \[m\text{vr=}\frac{h}{2\pi }\Rightarrow m\text{v=}\frac{h}{2\pi r}\] Now, \[m\text{v=}\frac{h}{\lambda }\] So, \[\frac{h}{\lambda }=\frac{h}{2\pi r}\Rightarrow \lambda =2\pi r\] \[\lambda =2\times 3.14\times 0.53\overset{o}{\mathop{A}}\,=3.328\overset{o}{\mathop{A}}\,\] \[=3.328\times {{10}^{-10}}m\] \[=0.3328\times {{10}^{-9}}m=0.3328nm\]
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