(solved)Question 2.34 of NCERT Class XI Chemistry Chapter 2

by Bohr's theory, IE is the energy
required to remove an electron from
the n-th shell of a H-like atom of
atomic number Z:

$\displaystyle E_n = +R_H\bigg(\frac{Z^2}{n^2}\bigg)$ J/atom

∴ IE of hydrogen for n = 1 and Z = 1 is

$\displaystyle E_1 = R_H \bigg(\frac{1^2}{1^2}\bigg) = R_H \text{ (J/atom)}$

but given $\displaystyle E_1 = 2.18 \times 10^{-18}\text{ J/atom}$

$\displaystyle \therefore R_H = 2.18 \times 10^{-18}\text{ . . . (1)}$

He⁺ is hydrogen-like with 1 electron

hence, IE for He⁺ (Z = 2, n = 1)

$\displaystyle \therefore \text{ IE } = +R_H\bigg(\frac{2^2}{1^2}\bigg) = R_H \times 4 $

$\displaystyle = 2.18\times10^{-18} \times 4 \:\bigg[\text{R}_H \text{ by (1)}\bigg]$

$\displaystyle = 8.72 \times 10^{-18} \text{ J/atom }\underline{Ans}$