A hydrogen like system has ionization energy 11808 kj/mol . Find the no. Of protons in the nucleus of the system
Answers
A hydrogen like system has ionisation energy 11808 kj/mol.
so, ionisation energy = 11808 × 10³ /(6.022 × 10²³) J/atom [ as we know, 1 mol = 6.023 × 10²³ ]
= (11808/6.022) × 10^-20 J/atom
= 1960.8 × 10^-20 J/atom
= 1.9608 × 10^-17 J/atom
we know, 1 eV/atom = 1.6 × 10^-19 J/atom
so, 1.9608 × 10^-17 J/atom = (1.9608/1.6) × 10² = 122.55 eV/atom
formula of ionisation energy , I.E = 13.6 × Z²/n² eV/atom
for first ionisation , n = 1
then, 122.55eV/atom = 13.6 × Z²/1
or, 9 = Z² => Z = 3
hence, number of protons in the nucleus of the system = 3.
Concept:
The basic minimum amount of energy needed to ionize an atom or an ion in a gaseous state. Kilojoules per mole (kJ/M) or electron volts are the two most commonly used measures of ionization energy (eV). On the periodic table, ionization energy displays periodicity.
Given:
The ionization energy = 11808 kJ/mol
Find:
Calculate the number of protons in the nucleus of the system.
Solution:
Ionization energy in a system similar to hydrogen is 11808 kj/mol.
Ionization energy is, therefore, equal to 11808 * 10³ / (6.022 * 10²³) J/atom [1 mol is equal to 6.023 * 10²³],
= (11808 / 6.022) * 10⁻²⁰ J/atom
= 1960.8 * 10⁻²⁰ J/atom
= 1.9608 * 10⁻¹⁷ J/atom
The known value of one eV per atom is 1.6 * 10⁻¹⁹ J/atom.
Thus, 1.9608 * 10⁻¹⁷ J/atom resulting in (1.9608 * 10⁻¹⁷ J/atom / 1.6 * 10⁻¹⁹ J/atom)
= (1.9608 / 1.6) * 10²
= 122.55 eV/atom
The formula of ionization energy,
I.E = 13.6 Z²/n² eV/atom
For first ionization, where n = 1 and, thus,
122.55 eV/atom = 13.6 Z²/ 1²
122.55 eV/atom / 13.6 = Z²
9 = Z²
√9 = Z
Z = 3
Hence, the number of protons in the nucleus of the system is 3.
The correct answer is 3.
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