Question

Bohrs radius of $2nd$ orbit of $B{e}^{3+}$ is equal to that of

• A. 4th orbit of hydrogen
• B. 2nd orbit of $H{e}^{+}$
• C. 3rd orbit of $L{i}^{2+}$
• D. first orbit of hydrogen

Answer 1

Answer: (D)

Bohrs radius for nth orbit $=0.53\times \frac{n^2}{Z}$
where, $Z=$ atomic number
$\therefore$ Bohrs radius of 2nd orbit of
$B{e}^{3+}=\frac{0.53\times (2{)}^2}{4}=0.53\mathring{A}$
(a) Bohrs radius of 4th orbit of
$H=\frac{0.53\times (4{)}^2}{1}=0.53\times 16\mathring{A}$
(b) Bohrs radius of 2nd orbit of
$H{e}^{+}=\frac{0.53\times (2{)}^2}{2}=0.53\times 2\mathring{A}$
(c) Bohrs radius of 3rd orbit of
$L{i}^{2+}=\frac{0.53\times (3{)}^2}{3}=0.53\times 3\mathring{A}$
(d) Bohrs radius of 1st orbit of
$H=\frac{0.53\times (1{)}^2}{1}=0.53\mathring{A}$
Hence, Bohrs radius of 2nd orbit of $B{e}^{3+}$ is equal to that of first orbit of hydrogen.