
An $\alpha $ particle is placed in an electric field at a point that has electric potential $5\,V$. Find its potential energy?
A. $1.6 \times {10^{ - 18}}\,J$
B. $1.6 \times {10^{ - 14}}\,J$
C. $1.6 \times {10^{ - 12}}\,J$
D. $1.6 \times {10^{ - 10}}\,J$
Answer
232.8k+ views
Hint In the question, the value of the electric potential is given. By using the expression of the potential energy and substituting the charge of the electron, then we get the value of the potential charge.
Formula used:
The expression for finding the potential energy is
${\text{Potential Energy = Potential}}\left( q \right){\text{ }} \times {\text{ charge}}\left( V \right)$
Where,
$q$ be the potential and $V$ be the charge
Complete step by step answer
Given that
Electric potential $ = \,5\,V$
Electric potential is the amount of work needed to move a unit of charge from a reference point to a specific point of the work. It is against an electric field. Generally, the reference point is the earth, whereas any point beyond the influence of the electric field charge. It is called the electric potential at a point in an electric field.
We know that the $\alpha $particle is placed in an electric field, so the charge of the alpha particle, we get
The charge of the $\alpha $particle $\left( q \right) = 2e$.
Energy of the electron $\left( e \right) = 1.6 \times {10^{ - 19}}$.
Substitute the value of the $e$in the expression of $q$, we get
$q = 2 \times 1.6 \times {10^{ - 19}}$
Performing the arithmetic operations in the above equation, we get
$q = 3.2 \times {10^{ - 19}}\,C$
Now, we have to find the potential energy, we get
${\text{Potential Energy = Potential}}\left( q \right){\text{ }} \times {\text{ charge}}\left( V \right)$
Substitute the known values in the above equation.
${\text{Potential Energy = }}3.2 \times {10^{ - 19}}\,C{\text{ }} \times {\text{ 5}}$
Performing the arithmetic operations, we get
${\text{Potential Energy}} = 1.6 \times {10^{ - 18}}\,J.$
Therefore, the potential energy is $1.6 \times {10^{ - 18}}\,J$.
Hence from the above options, option A is correct.
Note In the question, the alpha particle is placed in an electric field. Bu, an alpha particle is identical to a helium atom that has been stripped of its two of the electrons. So, the alpha particle has two protons and two neutrons. So, we substitute the values in the expression of potential energy.
Formula used:
The expression for finding the potential energy is
${\text{Potential Energy = Potential}}\left( q \right){\text{ }} \times {\text{ charge}}\left( V \right)$
Where,
$q$ be the potential and $V$ be the charge
Complete step by step answer
Given that
Electric potential $ = \,5\,V$
Electric potential is the amount of work needed to move a unit of charge from a reference point to a specific point of the work. It is against an electric field. Generally, the reference point is the earth, whereas any point beyond the influence of the electric field charge. It is called the electric potential at a point in an electric field.
We know that the $\alpha $particle is placed in an electric field, so the charge of the alpha particle, we get
The charge of the $\alpha $particle $\left( q \right) = 2e$.
Energy of the electron $\left( e \right) = 1.6 \times {10^{ - 19}}$.
Substitute the value of the $e$in the expression of $q$, we get
$q = 2 \times 1.6 \times {10^{ - 19}}$
Performing the arithmetic operations in the above equation, we get
$q = 3.2 \times {10^{ - 19}}\,C$
Now, we have to find the potential energy, we get
${\text{Potential Energy = Potential}}\left( q \right){\text{ }} \times {\text{ charge}}\left( V \right)$
Substitute the known values in the above equation.
${\text{Potential Energy = }}3.2 \times {10^{ - 19}}\,C{\text{ }} \times {\text{ 5}}$
Performing the arithmetic operations, we get
${\text{Potential Energy}} = 1.6 \times {10^{ - 18}}\,J.$
Therefore, the potential energy is $1.6 \times {10^{ - 18}}\,J$.
Hence from the above options, option A is correct.
Note In the question, the alpha particle is placed in an electric field. Bu, an alpha particle is identical to a helium atom that has been stripped of its two of the electrons. So, the alpha particle has two protons and two neutrons. So, we substitute the values in the expression of potential energy.
Recently Updated Pages
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 6 Shift 2 Question Paper with Answer Key

JEE Main 2023 (January 31 Evening Shift) Question Paper with Solutions [PDF]

JEE Main 2023 January 30 Shift 2 Question Paper with Answer Key

JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

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

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Understanding Uniform Acceleration in Physics

Understanding the Electric Field of a Uniformly Charged Ring

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Derivation of Equation of Trajectory Explained for Students

