Question

The radius of the Bohr orbit in the ground state of hydrogen atom is $0.5\mathring{\text{A}}$. The radius of the orbit of the electron in the third excited state of $He$ will be,

• A. $8\mathring{\text{A}}$
• B. $4\mathring{\text{A}}$
• C. $0.5\mathring{\text{A}}$
• D. $0.25\mathring{\text{A}}$

Answer 1

The correct option is B $4\mathring{\text{A}}$

As, $r_n=\frac{r_0{n}^2}{Z},\text{here,}{r}_0=0.5\mathring{\text{A}}$

Where $r_n$ is the radius of Bohr's orbit in the ground state atom.
In third excited state, $n=4$
And, for $He,Z=2$
$\therefore$ the radius of the orbit of the electron in the third excited state of $H{e}^{+}$ will be,
$r_4=\frac{r_0(4{)}^2}{2}$

$=\frac{0.5\times 16}{2}=4\mathring{\text{A}}$

Hence, $\left(B\right)$ is the correct answer.

Why this question ? This question is based on the radius of $n^{\text{th}}$ Bohr's orbit.