A) 4th orbit of hydrogen
B) 2nd orbit of\[H{{e}^{+}}\]
C) 3rd orbit of\[L{{i}^{2+}}\]
D) first orbit of hydrogen
Correct Answer: D
Solution :
Bohrs radius for nth orbit\[=0.53\times \frac{{{n}^{2}}}{Z}\] where, \[Z=\]atomic number \[\therefore \]Bohrs radius of 2nd orbit of \[B{{e}^{3+}}=\frac{0.53\times {{(2)}^{2}}}{4}=0.53\overset{\text{o}}{\mathop{\text{A}}}\,\] (a) Bohrs radius of 4th orbit of \[H=\frac{0.53\times {{(4)}^{2}}}{1}=0.53\times 16\overset{\text{o}}{\mathop{\text{A}}}\,\] (b) Bohrs radius of 2nd orbit of \[H{{e}^{+}}=\frac{0.53\times {{(2)}^{2}}}{2}=0.53\times 2\overset{\text{o}}{\mathop{\text{A}}}\,\] (c) Bohrs radius of 3rd orbit of \[L{{i}^{2+}}=\frac{0.53\times {{(3)}^{2}}}{3}=0.53\times 3\overset{\text{o}}{\mathop{\text{A}}}\,\] (d) Bohrs radius of 1st orbit of \[H=\frac{0.53\times {{(1)}^{2}}}{1}=0.53\overset{\text{o}}{\mathop{\text{A}}}\,\] Hence, Bohrs radius of 2nd orbit of\[B{{e}^{3+}}\]is equal to that of first orbit of hydrogen.
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