Question

Question 3.15 Energy of an electron in the ground state of the hydrogen atom is –2.18 × 10^(–18) J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol–1.

Class XI Classification of Elements and Periodicity in Properties Page 93

Answer 1

we know, ionisation energy is the amount of energy required to remove an electron from isolated gaseous atom in ground state.
e.g., I.E is equal to energy required to remove from E1 ( ground state) to E∞ ( infinity) .
$e.g., I.E=E_{\infty}-E_1$
E1 = -2.18 × 10^-18 J
$E_{ \infty}=0$
now, I.E = 0 - (-2.18 × 10^-18) = 2.18 × 10^-18 J
hence, I.E per H - atom = 2.18 × 10^-18 J
so, I.E per mole of H - atom = Avogadro's constant × 2.18 × 10^-18 J
= 6.022 × 10²³ × 2.18 × 10^-18 J
= 13.12 × 10^5 J

hence , Ionisation enthalpy = 13.12 × 10^5 J/mol

Answer 2

Answer:

we know, ionisation energy is the amount of energy required to remove an electron from isolated gaseous atom in ground state.

e.g., I.E is equal to energy required to remove from E1 ( ground state) to E∞ ( infinity) .

e.g., I.E=E_{\infty}-E_1e.g.,I.E=E∞−E1

E1 = -2.18 × 10^-18 J

E_{ \infty}=0E∞=0

now, I.E = 0 - (-2.18 × 10^-18) = 2.18 × 10^-18 J

hence, I.E per H - atom = 2.18 × 10^-18 J

so, I.E per mole of H - atom = Avogadro's constant × 2.18 × 10^-18 J

= 6.022 × 10²³ × 2.18 × 10^-18 J

= 13.12 × 10^5 J

hence , Ionisation enthalpy = 13.12 × 10^5 J/mol