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

The correct option is first orbit of He⁺.

According to Bohr’s model of the hydrogen-like atom, the kinetic energy (K.E.) of an electron in an orbit with principal quantum number n and atomic number Z is given by the formula:

$\text{K.E.}=13.6\times \frac{Z^2}{n^2}\text{eV}$

To find which state has the highest kinetic energy, we evaluate the value of $\frac{Z^2}{n^2}$ for each given option:

1. First orbit of H atom:
For Hydrogen (H), atomic number $Z=1$ and orbit number $n=1$.
$\frac{Z^2}{n^2}=\frac{1^2}{1^2}=1$

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

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

4. Second orbit of Li²⁺:
For Lithium ion (Li²⁺), atomic number $Z=3$ and orbit number $n=2$.
$\frac{Z^2}{n^2}=\frac{3^2}{2^2}=\frac{9}{4}=2.25$

Comparing all the calculated values of $\frac{Z^2}{n^2}$, the maximum ratio is 4, corresponding to the first orbit of He⁺. Therefore, the electron has the highest kinetic energy in the first orbit of He⁺.