Question 142: Atoms & Nuclei

Question

Considering the Bohr model of hydrogen like atoms, the ratio of the radius of $5^{	ext {th }}$ orbit of the electron in $\mathrm{Li}^{2+}$ and $\mathrm{He}^{+}$is

Accepted Answer

In the Bohr model for a hydrogen-like atom, the radius of the $ n^{	ext{th}} $ orbit is given by
$$ r_n = \frac{n^2 a_0}{Z} $$
where:
$ n $ is the principal quantum number,
$ a_0 $ is the Bohr radius, and
$ Z $ is the atomic number (the effective nuclear charge).
For the $ 5^{	ext{th}} $ orbit ($ n = 5 $):
For $ \mathrm{Li}^{2+} $ (where $ Z = 3 $):
$$    r_5(\mathrm{Li}^{2+}) = \frac{5^2 a_0}{3} = \frac{25a_0}{3}    $$
For $ \mathrm{He}^{+} $ (where $ Z = 2 $):
$$    r_5(\mathrm{He}^{+}) = \frac{5^2 a_0}{2} = \frac{25a_0}{2}    $$
Now, the ratio of the radius of the 5th orbit of $ \mathrm{Li}^{2+} $ to that of $ \mathrm{He}^{+} $ is:
$$ \frac{r_5(\mathrm{Li}^{2+})}{r_5(\mathrm{He}^{+})} = \frac{\frac{25a_0}{3}}{\frac{25a_0}{2}} = \frac{25a_0}{3} 	imes \frac{2}{25a_0} = \frac{2}{3} $$
Thus, the ratio is $ \frac{2}{3} $.
The correct answer is Option B.

Answer

$\frac{3}{2}$

Answer

$\frac{2}{3}$

Answer

$\frac{4}{9}$

Answer

$\frac{9}{4}$

Atoms & Nuclei

Considering the Bohr model of hydrogen like atoms, the ratio of the radius of 5^{\text {th }} orbit of the electron in \mathrm{Li}^{2+} and \mathrm{He}^{+}is

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