Question 334: Atoms & Nuclei

Question

The radius of fifth orbit of the $$\mathrm{Li}^{++}$$ is __________ $$	imes 10^{-12} \mathrm{~m}$$.
Take: radius of hydrogen atom $$ = 0.51\,\mathop A\limits^o $$

Accepted Answer

The formula to calculate the radius of an orbit for a hydrogen-like atom/ion is:
$$
r_n = r_0 \frac{n^2}{Z}
$$
where:
$r_n$ is the radius of the nth orbit,
$n$ is the principal quantum number (the orbit number),
$r_0$ is the Bohr radius (radius of the first Bohr orbit in the hydrogen atom), and
$Z$ is the atomic number (the number of protons in the nucleus).
We're dealing with a LiÃÂ²Ã¢ÂÂº ion and we're interested in the fifth orbit ($n = 5$), and given that $r_0$ is 0.51 ÃÂ and $Z$ for Li is 3, we can substitute these values into the formula:
$$
r_5 = 0.51 	imes \frac{25}{3} 	ext{ ÃÂ} = 4.25 	ext{ ÃÂ}
$$
which is $4.25 	imes 10^{-10}$ m, or equivalently $425 	imes 10^{-12}$ m when converted to meters.

Atoms & Nuclei

The radius of fifth orbit of the $\mathrm{Li}^{++} is __________ \times 10^{-12} \mathrm{~m}. Take: radius of hydrogen atom = 0.51\,\mathop A\limits^o

Integer / Numeric Answer Type