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# At which excited state of $B{e^{3 + }}$ radius of ${e^ - }$ will be the same as $H$ atoms and electrons in ground state.A) $1$B) $2$C) $3$D) $4$

Last updated date: 11th Sep 2024
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Hint: We can approach the solution to this question by simply substituting the value of atomic number $\left( Z \right)$ of Beryllium atom in the equation of radius of a Hydrogen-like atom. The equation is as follows:
$r\, = \,{r_o}\, \times \,\dfrac{{{n^2}}}{Z}$

We will be trying to solve the question exactly like we foretold in the hint section of the solution. Firstly, we should know the value of the atomic number of the Beryllium atom and the equation of radius of a Hydrogen-like atom which we told you in the hint section of the solution.
Let us have a look at what the question has given to us:
The atom is $B{e^{3 + }}$ , i.e. Beryllium with a $+ 3$ charge, which signifies that there is only one electron in the atom. This makes it a Hydrogen-like atom and enables us to use the formula in this situation.
Other than that, question has told us that it is a Beryllium atom, which tells us that the value of atomic number that we need to use is, $Z\, = \,4$
Now, all we need to do is to use the formula or, the equation of radius of a Hydrogen-like atom here.
Let’s have a look at the equation again:
$r\, = \,{r_o}\, \times \,\dfrac{{{n^2}}}{Z}$
The question asks us about the excited-state at which the radius $\left( r \right)$ becomes equal to the radius of Hydrogen atom in ground state, which is ${r_o}$ .
Putting this in the equation, we get:
${r_o}\, = \,{r_o}\, \times \,\dfrac{{{n^2}}}{Z}$
Which we can further simplified as:
${n^2}\, = \,Z$
We already discussed the value of the atomic number that we will be using, so:
${n^2}\, = \,4$
Taking square root both sides, we get:
$n\, = \,2$
Here we get the value of $n\, = \,2$ , which is the first excited state.

So, the correct option is option (A).

Note: Many students use the formula of Radius of an atom instead of radius of a Hydrogen-like atom which takes them to a wrong answer. Other than that, many students take $n\, = \,2$ as the second excited state instead of first and tick the wrong option which causes them to lose marks.