A) 0.053 nm
B) 0.106 nm
C) 0.2116 nm
D) 0.4256 nm
Correct Answer: C
Solution :
Given that: \[R=8.314Joul\,{{K}^{-1}}mo{{l}^{-1}}\] Bohrs orbit of hydrogen atom\[(n)=2\] Atomic number of hydrogen\[(z)=1\] By using: \[r=\frac{0.529{{n}^{2}}}{z}\] \[=\frac{0.529\times {{(2)}^{2}}}{1}\] \[=\frac{0.529\times 4}{1}\] \[=2.116\overset{o}{\mathop{\text{A}}}\,\] \[=0.2116\text{ }nm\]
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