Advertisements
Advertisements
प्रश्न
Energy of an electron in the ground state of the hydrogen atom is −2.18 × 10−18 J. Calculate the ionisation enthalpy of atomic hydrogen in terms of kJ mol−1.
Advertisements
उत्तर
Energy of an electron in the ground state of the hydrogen atom = −2.18 × 10−18 J
\[\ce{H -> H^+ + e^-}\]
The energy required to ionize 1 mole of hydrogen atoms, we multiply by the Avogadro constant.
E = 2.18 × 10−18 × 6.023 × 1023
= 13.123 × 105 J mol−1
I.E = +1312 K J mol−1
APPEARS IN
संबंधित प्रश्न
Which one of the following is the least electronegative element?
The First ionisation potential of Na, Mg and Si are 496, 737 and 786 kJ mol-1 respectively. The ionisation potential of Al will be closer to
What are isoelectronic ions? Give examples.
What is the effective nuclear charge?
Magnesium loses electrons successively to form Mg+, Mg2+ and Mg3+ ions. Which step will have the highest ionisation energy and why?
How would you explain the fact that the second ionisation potential is always higher than the first ionisation potential?
Explain the Pauling method for the determination of ionic radius.
By using Pauling's method calculate the ionic radii of K+ and Cl− ions in the potassium chloride crystal. Given that `"d"_("K"^+) - "Cl"^-` = 3.14 Å
Explain the following, give an appropriate reason.
The formation of \[\ce{F^-_{(g)}}\] from \[\ce{F_{(g)}}\] is exothermic while that of \[\ce{O^2-_{(g)}}\] from \[\ce{O_{(g)}}\] is endothermic.
State the trends in the variation of electronegativity in groups and periods.
