Question

Which level of the doubly-ionised lithium $\left(L{i}^{++}\right)$, which $Z=3$, has the same energy as the ground state energy of hydrogen atom $(Z=1)$. Compare the orbital radii of the two levels.

Answer 1

The energy of an atom at nth state

${ϵ}_n={ϵ}_0\frac{z^2}{n^2}$

Given that,

Let nth level of $L{i}^{++}$ has the same energy as that of ground state of Hydrogen atom

as $z=3$ for $Li$

$\Rightarrow \left({ϵ}_n\right)$ of $Li=\left({ϵ}_1\right)$ of $H$

$\Rightarrow {ϵ}_0\frac{(3{)}^2}{n^2}={ϵ}_0\frac{(1{)}^2}{(1{)}^2}$

$\Rightarrow n=3$

Now, 3rd level of $L{i}^{++}$ has same energy as 1st level (Ground state) of Hydrogen radius of nth level $\left(r_n\right)=r_0\frac{n^2}{z}$

$\Rightarrow \frac{r_3\left(Li\right)}{r_1\left(H\right)}=\frac{r_0\frac{(3{)}^2}{3}}{r_0\left(\frac{(1{)}^2}{1}\right)}=3$

the radius of 3rd level of $L{i}^{++}$ is $3$ times the radius of ground state of hydrogen.