Ionisation energy of He+ is 19.6 × 10-18 J atom-1. The energy of the first stationary state (n= 1) of Li2+ is

Question

Ionisation energy of He+ is 19.6 × 10-18 J atom-1. The energy of the first stationary state (n= 1) of Li2+ is (A) 4.41 × 10-16 J atom-1 (B) -4.41 × 10-17 J atom-1 (C) -2.2 × 10-15 J atom-1 (D) 8.82 × 10-17 J atom-1

Tardigrade Admin · Accepted Answer

I.E= (Z2/n2)×13.6 eV...(i) or (I1/I2) = (Z21/n21)×(n22/Z22) ...(ii) Given I1 = -19.6 × 10-18, Z1 = 2, n1 = 1, Z2 = 3 and n2 = 1 Substituting these values in equation (ii). -(19.6×10-18/I2) = (4/1)×(1/9) or I2=19.6× 10-18× (9/4) = -4.41 × 10-17 J/atom

Q. Ionisation energy of $H e^{+}$ is $19.6 \times 1 0^{- 18} J$ atom $^{- 1}$ . The energy of the first stationary state $(n = 1)$ of $L i^{2 +}$ is

$4.41 \times 1 0^{- 16} J a t o m^{- 1}$

$- 4.41 \times 1 0^{- 17} J a t o m^{- 1}$

$- 2.2 \times 1 0^{- 15} J a t o m^{- 1}$

$8.82 \times 1 0^{- 17} J a t o m^{- 1}$

Q. Ionisation energy of He+ is 19.6×10−18J atom−1. The energy of the first stationary state (n=1) of Li2+ is

4.41×10−16Jatom−1

−4.41×10−17Jatom−1

−2.2×10−15Jatom−1

8.82×10−17Jatom−1

I.E=n2Z2​×13.6eV...(i) or I2​I1​​=n12​Z12​​×Z22​n22​​...(ii) Given I1​=−19.6×10−18,Z1​=2, n1​=1,Z2​=3 and n2​=1 Substituting these values in equation (ii). −I2​19.6×10−18​=14​×91​ or I2​=19.6×10−18×49​ =−4.41×10−17J/atom

Q. Ionisation energy of $H e^{+}$ is $19.6 \times 1 0^{- 18} J$ atom $^{- 1}$ . The energy of the first stationary state $(n = 1)$ of $L i^{2 +}$ is

$4.41 \times 1 0^{- 16} J a t o m^{- 1}$

$- 4.41 \times 1 0^{- 17} J a t o m^{- 1}$

$- 2.2 \times 1 0^{- 15} J a t o m^{- 1}$

$8.82 \times 1 0^{- 17} J a t o m^{- 1}$