JEE MAIN - Chemistry (2010 - No. 25)
Ionisation energy of He+ is 19.6 x 10–18 J atom–1. The energy of the first stationary state (n = 1) of Li2+ is
4.41 x 10–16 J atom–1
-4.41 x 10–17 J atom–1
-2.2 x 10–15 J atom–1
8.82 x 10–17 J atom–1
Explanation
Ionisation energy(IE) - It is the energy required to move an electron from ground state to infinity.
IE = $${E_\infty } - {E_1}$$ = $$0 - {E_1}$$ = $$ - {E_1}$$
$$\therefore$$ E1 of He+ = - 19.6 x 10–18 J atom–1
Energy of a species at n state,
(En)species = (En)hydrogen $$ \times $$ Z2
$$\therefore$$ (E1)hydrogen = $${{ - 19.6 \times {{10}^{ - 18}}} \over 4}$$ [ For He, Z = 2 ]
(E1)Li+2 = $${{ - 19.6 \times {{10}^{ - 18}}} \over 4} \times {3^2}$$
= -4.41 x 10–17 J atom–1
IE = $${E_\infty } - {E_1}$$ = $$0 - {E_1}$$ = $$ - {E_1}$$
$$\therefore$$ E1 of He+ = - 19.6 x 10–18 J atom–1
Energy of a species at n state,
(En)species = (En)hydrogen $$ \times $$ Z2
$$\therefore$$ (E1)hydrogen = $${{ - 19.6 \times {{10}^{ - 18}}} \over 4}$$ [ For He, Z = 2 ]
(E1)Li+2 = $${{ - 19.6 \times {{10}^{ - 18}}} \over 4} \times {3^2}$$
= -4.41 x 10–17 J atom–1
Comments (0)
