According to Bohr’s model, the highest kinetic energy is associated with the electron in the

To determine which orbit and species has the highest kinetic energy associated with its electron, we can use Bohr's model of the hydrogen-like atoms.

According to Bohr's model, the total energy $E$ of an electron in the $n$-th orbit of a species with atomic number $Z$ is given by the formula:
$E=-13.6\times \frac{Z^2}{n^2}\text{eV}$

The relationship between the total energy ($E$), potential energy ($U$), and kinetic energy ($K$) of the electron is:
$K=-E$
Therefore, the kinetic energy of the electron in the $n$-th orbit is:
$K=13.6\times \frac{Z^2}{n^2}\text{eV}$

This means the kinetic energy is directly proportional to $\frac{Z^2}{n^2}$. Let us calculate the value of $\frac{Z^2}{n^2}$ for each of the given options:

1. first orbit of H atom:
For Hydrogen (H), $Z=1$ and for the first orbit, $n=1$.
$\frac{Z^2}{n^2}=\frac{1^2}{1^2}=1$

2. first orbit of He⁺:
For Helium ion (He⁺), $Z=2$ and for the first orbit, $n=1$.
$\frac{Z^2}{n^2}=\frac{2^2}{1^2}=4$

3. second orbit of He⁺:
For Helium ion (He⁺), $Z=2$ and for the second orbit, $n=2$.
$\frac{Z^2}{n^2}=\frac{2^2}{2^2}=1$

4. second orbit of Li²⁺:
For Lithium ion (Li²⁺), $Z=3$ and for the second orbit, $n=2$.
$\frac{Z^2}{n^2}=\frac{3^2}{2^2}=2.25$

Comparing the values, the ratio $\frac{Z^2}{n^2}$ is highest for the first orbit of He⁺ (value of 4), which corresponds to the maximum kinetic energy:
$K=13.6\times 4=54.4\text{eV}$

Thus, the highest kinetic energy is associated with the electron in the first orbit of He⁺.