
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
135.3k+ 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.
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