A) \[0.176\text{ }\overset{o}{\mathop{\text{A}}}\,\]
B) \[\text{0}\text{.30 }\overset{o}{\mathop{\text{A}}}\,\]
C) \[\text{0}\text{.53 }\overset{o}{\mathop{\text{A}}}\,\]
D) \[\text{1}\text{.23 }\overset{o}{\mathop{\text{A}}}\,\]
Correct Answer: A
Solution :
Key Idea \[{{r}_{n}}=\frac{{{r}_{0}}\times {{n}^{2}}}{Z}\] Given,\[{{r}_{0}}=\]radius of H atom in ground state \[=0.5\overset{o}{\mathop{\text{A}}}\,\] \[n=\]number of orbit = 1 \[Z=\]Atomic number of\[Li=3\] \[\therefore \] \[{{r}_{n}}=\frac{0.53\times {{1}^{2}}}{3}=0.176\overset{o}{\mathop{\text{A}}}\,\]
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