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Q3.15:
Energy of an electron in the ground state of the hydrogen atom is (–2.18 \times 10^{-18}J). Calculate the ionization enthalpy of atomic hydrogen in terms of (J mol^{-1}). Hint: Apply the idea of mole concept to derive the answer.

Solution :

Step 1: Defining the Theoretical Basis

The ionization enthalpy of a hydrogen atom is defined as the minimum amount of energy required to remove an electron from the ground state ($n=1$) to infinity ($n=\infty$), where the electron is no longer bound to the nucleus. The energy of the electron at $n=\infty$ is defined as $0 \text{ J}$.

The energy required for the ionization of a single hydrogen atom ($\Delta E_{atom}$) is calculated as follows:

$\Delta E_{atom} = E_{\infty} - E_{ground}$

Given: $E_{ground} = -2.18 \times 10^{-18} \text{ J}$

$\Delta E_{atom} = 0 - (-2.18 \times 10^{-18} \text{ J}) = +2.18 \times 10^{-18} \text{ J}$

[Per the Bohr Model of the Hydrogen Atom]

Step 2: Application of the Mole Concept

The ionization enthalpy requested is per mole ($J \text{ mol}^{-1}$). Since one mole of any substance contains Avogadro’s constant ($N_A$) number of particles, we must multiply the energy required for one atom by $N_A$.

Avogadro's Constant ($N_A$) $\approx 6.022 \times 10^{23} \text{ mol}^{-1}$

Formula: $\Delta H_{ionization} = \Delta E_{atom} \times N_A$

Step 3: Algebraic Calculation

Substitute the values into the equation:

$\Delta H_{ionization} = (2.18 \times 10^{-18} \text{ J/atom}) \times (6.022 \times 10^{23} \text{ atoms/mol})$

$\Delta H_{ionization} = (2.18 \times 6.022) \times 10^{(-18 + 23)} \text{ J mol}^{-1}$

$\Delta H_{ionization} = 13.12796 \times 10^5 \text{ J mol}^{-1}$

[By applying the laws of exponents and arithmetic multiplication]

Step 4: Expressing in Standard Scientific Notation

Adjusting the decimal to adhere to standard scientific notation:

$\Delta H_{ionization} = 1.312796 \times 10^6 \text{ J mol}^{-1}$

Rounding to three significant figures consistent with the provided data ($2.18$):

$\Delta H_{ionization} \approx 1.31 \times 10^6 \text{ J mol}^{-1}$

Ground State (n=1): -2.18 x 10^-18 J Ionized State (n=∞): 0 J ΔE = +2.18 x 10^-18 J

Final Solution: The ionization enthalpy of atomic hydrogen is $1.31 \times 10^6 \text{ J mol}^{-1}$.


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