A) \[4\pi {{a}_{0}}\]
B) \[6\pi {{a}_{0}}\]
C) \[8\pi {{a}_{0}}\]
D) \[2\pi {{a}_{0}}\]
Correct Answer: C
Solution :
[c] According to Bohr's model \[{{r}_{n}}=\frac{{{n}^{2}}}{Z}\times {{a}_{0}}\] (\[{{a}_{0}}={{1}^{st}}\]Bohr radius) \[\because 2\pi r=n\lambda \] (using de-Broglie relation) \[\Rightarrow 2\pi \times \frac{{{4}^{2}}}{1}\times {{a}_{0}}=4\lambda \] \[\Rightarrow \lambda =8\pi {{a}_{0}}\]
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