A) \[2\pi {{a}_{0}}\]
B) \[16{{a}_{0}}\]
C) \[{{a}_{0}}/4\]
D) \[8\pi {{a}_{0}}\]
Correct Answer: D
Solution :
[d]: \[{{r}_{n}}={{a}_{0}}\times {{n}^{2}},{{r}_{4}}={{a}_{0}}\times {{4}^{2}}=16{{a}_{0}}\] \[mv=4\times \frac{h}{2\pi {{r}_{4}}}=\frac{4h}{2\pi \times 16{{a}_{0}}}=\frac{h}{8\pi {{a}_{0}}}\] \[\therefore \]\[\lambda =\frac{h}{mv}=8\pi {{a}_{0}}\]
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