
The ionisation potential of a Helium atom is $24.6V$. The energy required to ionise it will be,
A) $24.6eV$
B) $34.6eV$
C) $30.6eV$
D) $13.6eV$
Answer
213.6k+ views
Hint: Recall what you understand by ionization energy. How it is affected by other factors and how the ionisation potential can be used to calculate the energy required to ionize an atom. Answers to these questions will help you solve these kinds of problems.
Complete step by step answer:
Let’s start with the definition of ionization energy.
Ionization energy of an atom is the minimum amount of energy required to knock out one electron from its outermost orbit or shell. In other terms, the ionisation energy is a measure of binding strength of the electrons to the nucleus.
The more is the ionisation energy the more will be the binding strength of the electron to the nucleus.
This is affected by following conditions.
If the atomic radii of the atom is higher, lower will be the ionisation energy because the outermost electron becomes loosely attached.
With increase in effective nuclear charge, the ionisation energy of the atom increases.
The ionization potential is the measure of the potential by which we can remove an electron from an atom. The work done by this potential is given by electron-Volts ($eV$).
For example, if the ionization potential of an atom is $V$ then the energy required to knock out an electron from the outermost shell of that atom is given by
$E = eV$
Where $e$ is the charge of an electron that is $1.6 \times {10^{ - 19}}C$.
This is because we know that energy required to move a charge \[q\] in an area where the potential is $V$, will be,
$E = qV$
Now for the given question, the ionisation potential for Helium atom is $24.6V$. This implies that an electron will be knocked out if we provide an energy equal to amount
$E = 24.6V \times 1e$
$\therefore E = 24.6eV$
Thus the correct answer will be option A, $24.6eV$.
Note: One can easily answer these kinds of questions because the ionisation energy is numerically equal to ionisation potential, but has the units of energy, that is, electron-Volts. Sometimes the examiner might ask the question where the unit of ionisation energy will be in $Joules$. In such a case you have to remember that, $1eV = 1.6 \times {10^{ - 19}}J$, and put this value in the answer.
Complete step by step answer:
Let’s start with the definition of ionization energy.
Ionization energy of an atom is the minimum amount of energy required to knock out one electron from its outermost orbit or shell. In other terms, the ionisation energy is a measure of binding strength of the electrons to the nucleus.
The more is the ionisation energy the more will be the binding strength of the electron to the nucleus.
This is affected by following conditions.
If the atomic radii of the atom is higher, lower will be the ionisation energy because the outermost electron becomes loosely attached.
With increase in effective nuclear charge, the ionisation energy of the atom increases.
The ionization potential is the measure of the potential by which we can remove an electron from an atom. The work done by this potential is given by electron-Volts ($eV$).
For example, if the ionization potential of an atom is $V$ then the energy required to knock out an electron from the outermost shell of that atom is given by
$E = eV$
Where $e$ is the charge of an electron that is $1.6 \times {10^{ - 19}}C$.
This is because we know that energy required to move a charge \[q\] in an area where the potential is $V$, will be,
$E = qV$
Now for the given question, the ionisation potential for Helium atom is $24.6V$. This implies that an electron will be knocked out if we provide an energy equal to amount
$E = 24.6V \times 1e$
$\therefore E = 24.6eV$
Thus the correct answer will be option A, $24.6eV$.
Note: One can easily answer these kinds of questions because the ionisation energy is numerically equal to ionisation potential, but has the units of energy, that is, electron-Volts. Sometimes the examiner might ask the question where the unit of ionisation energy will be in $Joules$. In such a case you have to remember that, $1eV = 1.6 \times {10^{ - 19}}J$, and put this value in the answer.
Recently Updated Pages
Chemical Equation - Important Concepts and Tips for JEE

JEE Main 2022 (July 29th Shift 1) Chemistry Question Paper with Answer Key

Conduction, Transfer of Energy Important Concepts and Tips for JEE

JEE Analytical Method of Vector Addition Important Concepts and Tips

Atomic Size - Important Concepts and Tips for JEE

JEE Main 2022 (June 29th Shift 1) Maths Question Paper with Answer Key

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

JEE Main Correction Window 2026 Session 1 Dates Announced - Edit Form Details, Dates and Link

Equation of Trajectory in Projectile Motion: Derivation & Proof

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Hybridisation in Chemistry – Concept, Types & Applications

Angle of Deviation in a Prism – Formula, Diagram & Applications

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Collision: Meaning, Types & Examples in Physics

Atomic Structure: Definition, Models, and Examples

Degree of Dissociation: Meaning, Formula, Calculation & Uses

How to Convert a Galvanometer into an Ammeter or Voltmeter

Average and RMS Value in Physics: Formula, Comparison & Application

