Question 50: Energy and radius of first Bohr orbit of He+ and Li2+ are
[Given RH= 2.18 x 10-18J, a0 = 52.9 pm]
(1) En (Li2+) =-19.62 x 10-18J
rn (Li2+) = 17.6 pm
En (He+) = -8.72 x 10-18J
rn (He+) = 26.4 pm
(2) En (Li2+) = -8.72 x 10-18J
rn (Li2+) = 26.4 pm
En (He+) = -19.62 x 10-18J
rn (He+) = 17.6 pm
(3) En (Li2+) = –19.62 x 10-16J
rn (Li2+) = 17.6 pm
En (He+) = -8.72 x 10-16J
rn (He+) = 26.4 pm
(4) En (Li2+) = -8.72 x 10-16J
rn (Li2+) = 17.6 pm
En (He2+) = -19.62 x 10-16J
rn (He+) = 17.6 pm
Answer: option(1)
Explanation:
For hydrogen-like ions (single electron systems such as He⁺ and Li²⁺),
Bohr model formulas are:
Energy of nth orbit:
E_n = -\frac{Z^2 R_H}{n^2}Radius of nth orbit:
r_n = \frac{n^2 a_0}{Z}Given:
R_H = 2.18 \times 10^{-18}\ \text{J} a_0 = 52.9\ \text{pm}First orbit ⇒ n = 1
For Li²⁺ (Z = 3):
Energy:
E_1(\text{Li}^{2+}) = -Z^2 R_H = -9 \times 2.18 \times 10^{-18} E_1(\text{Li}^{2+}) = -19.62 \times 10^{-18}\ \text{J}Radius:
r_1(\text{Li}^{2+}) = \frac{a_0}{Z} = \frac{52.9}{3} r_1(\text{Li}^{2+}) \approx 17.6\ \text{pm}For He⁺ (Z = 2):
Energy:
E_1(\text{He}^+) = -Z^2 R_H = -4 \times 2.18 \times 10^{-18} E_1(\text{He}^+) = -8.72 \times 10^{-18}\ \text{J}Radius:
r_1(\text{He}^+) = \frac{a_0}{Z} = \frac{52.9}{2} r_1(\text{He}^+) = 26.4\ \text{pm}Thus,
Li²⁺ → E_1 = -19.62 \times 10^{-18}\ \text{J},\ r_1 = 17.6\ \text{pm}
He⁺ → E_1 = -8.72 \times 10^{-18}\ \text{J},\ r_1 = 26.4\ \text{pm}
Hence, option (1) is correct.