Question

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

• A. 4th orbit of hydrogen
• B. 2nd orbit of $ He^{+}$
• C. 3rd orbit of $ Li^{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
$Be^{3+}=\frac{0.53 \times (2)^{2}}{4}=0.53\, \mathring{A}$
(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
$He^{+}=\frac{0.53 \times (2)^{2}}{2}=0.53 \times 2\,\mathring{A}$
(c) Bohrs radius of 3rd orbit of
$Li^{2+}=\frac{0.53 \times (3)^{2}}{3}=0.53 \times 3\,\mathring{A}$
(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 $ Be^{3+}$ is equal to that of first orbit of hydrogen.