Question

What is the radius of Bohr’s fifth orbit for ${B^{ + 4}}$ .

Answer 1

Bohr model of the hydrogen atom was the first atomic model to successfully explain the radiation spectra of the atomic hydrogen. The equation for the radius of the Bohr’s ${n^{th}}$ orbit is found using the atomic number, the number of orbits and the bohr’s radius. Bohr radius is a physical constant which expresses the most probable distance between the electron and the nucleus in a hydrogen atom in the ground state.

Complete step by step solution:
The equation to find the atomic radius of the ${n^{th}}$ Bohr orbit is
$r(n) = \dfrac{{{n^2}}}{Z} \times {r_0}$
Where,
‘n’ is a positive integer
‘Z’ is the atomic number
‘ ${r_0}$ ’ is the smallest allowed radius for the hydrogen atom which is also known as the Bohr’s radius.
The Bohr’s radius has a value of $0.529 \times {10^{ - 10}}m$
Since we have to find out the radius of the Bohr’s fifth orbit for ${B^{ + 4}}$ .
$n = 5$
Since we know that the atomic number of Boron is $5$ .
Hence, $Z = 5$
\ {r_5}({B^{ + 4}}) = \dfrac{{{n^2}}}{Z} \times {r_0} \\\ {r_5}({B^{ + 4}}) = \dfrac{{{5^2}}}{5} \times 0.529 \times {10^{ - 10}}m \\\ {r_5}({B^{ + 4}}) = \dfrac{{25}}{5} \times 0.529 \times {10^{ - 10}}m \\\ {r_5}({B^{ + 4}}) = 2.645 \times {10^{ - 10}}m \\\ \
Therefore, the radius of Bohr’s fifth orbit for ${B^{ + 4}}$ is $2.645 \times {10^{ - 10}}$ m.

Note:
Although the Bohr model is not that frequently used in physics, the Bohr radius is highly used due to its presence in calculating other fundamental physical constants for example like atomic units and fine structure constants. The electrons are orbiting around the nucleus in circular paths. Neils Bohr proposed that the electrons are surrounded and roam around the nucleus in an orbicular probability zone like shells. The importance of Bohr radius is very high. The smallest average radius can be attained by a neutral atom.