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Bohr's radius of $2 \mathrm{nd}$ orbit of $\mathrm{Be}^{3+}$ is equal to that of
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1st orbit of hydrogen
Bohr radius for $n$th orbit $=0.53 Å \times \frac{n^2}{Z}$
where, $Z=$ atomic number
$\therefore$ Bohr radius of 2 nd orbit of
$\mathrm{Be}^{3+}=\frac{0.53 \times(2)^3}{4}=0.53 Å$
For option (d) Bohr radius of 1st orbit of
$\mathrm{H}=\frac{0.53 \times(1)^3}{1}=0.53 Å$
Hence, Bohr's radius of $2 \mathrm{nd}$ orbit of $\mathrm{Be}^{3+}$ is equal to that of first orbit of hydrogen.
where, $Z=$ atomic number
$\therefore$ Bohr radius of 2 nd orbit of
$\mathrm{Be}^{3+}=\frac{0.53 \times(2)^3}{4}=0.53 Å$
For option (d) Bohr radius of 1st orbit of
$\mathrm{H}=\frac{0.53 \times(1)^3}{1}=0.53 Å$
Hence, Bohr's radius of $2 \mathrm{nd}$ orbit of $\mathrm{Be}^{3+}$ is equal to that of first orbit of hydrogen.
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