Energy and radius of first Bohr orbit of He⁺ and Li²⁺ are
[Given RH = 2.18 × 10^–18 J, a0 = 52.9 pm]

The correct option is:
E_n(Li²⁺) = -19.62 × 10^-18 J; r_n(Li²⁺) = 17.6 pm; E_n(He⁺) = -8.72 × 10^-18 J; r_n(He⁺) = 26.4 pm

Step-by-Step Derivation and Logic:

According to Bohr's model for hydrogen-like single-electron species (such as He⁺ and Li²⁺), the energy of an electron in the n-th orbit is given by the formula:

$E_n=-R_H\times \frac{Z^2}{n^2}$
where:
• R_H = 2.18 × 10^-18 J (Rydberg constant)
• Z is the atomic number of the element
• n is the principal quantum number (orbit number)

Similarly, the radius of the n-th Bohr orbit is given by:

$r_n=a_0\times \frac{n^2}{Z}$
where:
• a₀ = 52.9 pm (Bohr radius for hydrogen)

For the first Bohr orbit, we set n = 1.

1. Calculations for Helium Ion (He⁺):
For He⁺, the atomic number Z = 2.
• Energy of the first Bohr orbit (n = 1):

$E_1\left({\mathrm{He}}^{+}\right)=-(2.18\times {\mathrm{10}}^{-18}J)\times \frac{2^2}{1^2}=-2.18\times {\mathrm{10}}^{-18}\times 4=-8.72\times {\mathrm{10}}^{-18}J$
• Radius of the first Bohr orbit (n = 1):

$r_1\left({\mathrm{He}}^{+}\right)=52.9pm\times \frac{1^2}{2}=26.45pm\approx 26.4pm$

2. Calculations for Lithium Ion (Li²⁺):
For Li²⁺, the atomic number Z = 3.
• Energy of the first Bohr orbit (n = 1):


• Radius of the first Bohr orbit (n = 1):

$r_1\left({\mathrm{Li}}^{2+}\right)=52.9pm\times \frac{1^2}{3}=17.63pm\approx 17.6pm$