Question

If ${{a}_{0}}$​ is the Bohr radius, then the radius $n=2$ in triply ionized beryllium will be,
$\begin{align}
  & A.4{{a}_{0}} \\
 & B.{{a}_{0}} \\
 & C.\dfrac{{{a}_{0}}}{4} \\
 & D.2{{a}_{0}} \\
\end{align}$

Answer 1

Hint: The radius of the hydrogen equivalent system can be found by taking the ratio of the product of the square of the number of shells present and the radius of Bohr orbit to the atomic number of the atom. Here it is mentioned that the atom is a triply ionised beryllium. Triply ionised beryllium is having the number of electrons as one and atomic number as four. Substitute all this in the equation for the radius of the atom. This will help you in answering this question.

Complete step by step answer:
The radius of an hydrogen equivalent system which is having an atomic number mentioned as $Z$, is given to be as ${{r}_{n}}$. This can be written as,
${{r}_{n}}=\dfrac{{{n}^{2}}{{a}_{0}}}{Z}$
Where ${{a}_{0}}$be the radius of the Bohr orbit.
Here it is mentioned that the atom is a triply ionised beryllium. Triply ionised beryllium is having the number of electrons as one. The atomic number can be written as,
$Z=4$
And the number of orbit has been mentioned as,
$n=2$
Let us substitute this in the above cited equation. This can be written as,
\[{{r}_{n}}=\dfrac{{{2}^{2}}\times {{a}_{0}}}{4}={{a}_{0}}\]
Therefore the radius of the orbit has been calculated.
This has been given as option B.

Note:
The Bohr radius is basically defined as a physical constant. T will be equivalent to the most probable distance between the nucleus of an atom and the first orbital of the hydrogen where the electron revolves. It is referred to as the ground state of an atom which is non-relativistic and also with an infinitely heavy proton. It is named after the renowned scientist Neils Bohr. He explained the Bohr model of an atom.