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# If ${a_0}$ is the Bohr radius, the radius of then $n = 2$ electronics orbit in triply ionized beryllium is:A. $4{a_0}$B. ${a_0}$C. ${a_0}/4$D. ${a_0}/16$

Last updated date: 15th Aug 2024
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Hint: We know that atoms of different elements have different types of radius. There are many types of atomic radius such as a Van der Waals radius, Ionic radius, Covalent radius, Metallic radius and Bohr radius which explain the radius of atoms. We can solve this problem with the Bohr concept.

${r_n} = \dfrac{{{n^2}r}}{Z}$
Where, $Z$ is the atomic number of elements , $n$ is the number of orbits and $r$ is the total radius .
Applying the above formula in the given question we find that $r = {a_0}$and $n = 2$. The atomic no. for beryllium is 4 so the value of $Z$ is equal to $4$.
${r_n} = \dfrac{{{{\left( 2 \right)}^2}{a_0}}}{4}$ $\Rightarrow {a_0}$
Note: According to Bohr concept, Bohr radius is the most probable distance between nucleus and the electron in a hydrogen atom in its ground states. The value of the Bohr radius is $5.29177 \times {10^{ - 11}}m$.