The KE of the electron in an orbit of radius ‘r’ in hydrogen atom is proportional to (e = electronic charge)
$\begin{align}
& A.\,\,\,\dfrac{{{e}^{2}}}{{{r}^{2}}} \\
& B.\,\,\,\dfrac{{{e}^{2}}}{2{{r}^{3}}} \\
& C.\,\,\,\dfrac{{{e}^{2}}}{r} \\
& D.\,\,\,\dfrac{{{e}^{2}}}{2{{r}^{2}}} \\
\end{align}$
Answer
576.3k+ views
Hint: To get the value for kinetic energy, first get the value of potential energy and then with the relation between potential energy and kinetic energy, it will be easy to get the dependency of kinetic energy. The value for potential energy can be found using the predefined formula that is available.
Formula used:
$\begin{align}
& \text{Potential Energy (}P.E.)=\dfrac{k{{q}^{2}}}{r},\,\text{and} \\
& \text{Kinetic}\,\text{Energy}\,\text{(}K.E.)=\dfrac{1}{2}\left| P.E. \right| \\
\end{align}$
Complete solution Step-by-Step:
The formula of potential energy is:
$\text{Potential Energy (}P.E.)=\dfrac{k{{q}^{2}}}{r}$, where $k$ is a constant value.
Now, we know that potential energy is negative for electrons and $q=e$, as given in the question. So, the potential energy of electron in hydrogen atom with charge $e$ and radius $r$ is:
$\text{Potential Energy (}P.E.)=\dfrac{-k{{e}^{2}}}{r}$
Since, potential energy of the electron in hydrogen atom is deduced above, now we can get the kinetic energy of electron in hydrogen atom using the formula mentioned below:
$\text{Kinetic}\,\text{Energy}\,\text{(}K.E.)=\dfrac{1}{2}\left| P.E. \right|$
Substituting the value of potential energy in the above formula, we get:
$\begin{align}
& K.E.=\dfrac{1}{2}\left| P.E. \right| \\
& \Rightarrow K.E.=\dfrac{1}{2}\left| \dfrac{-k{{e}^{2}}}{r} \right| \\
& \Rightarrow K.E.=\dfrac{1}{2}\times \dfrac{k{{e}^{2}}}{r} \\
& \therefore K.E.=\dfrac{k{{e}^{2}}}{2r} \\
\end{align}$
From the above deduced value,
$K.E.\propto \dfrac{{{e}^{2}}}{r}$
Therefore, the correct answer is Option (C).
Additional Information:
In physics, the kinetic energy (KE) of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body when decelerating from its current speed to a state of rest.
Note:
This is not the only way possible to solve this type of problems. There are different ways in which the answer can be achieved and feel free to use the way you are comfortable, but remember the formula for whichever way you choose, as if the formula is now known, then getting the correct answer becomes difficult.
Formula used:
$\begin{align}
& \text{Potential Energy (}P.E.)=\dfrac{k{{q}^{2}}}{r},\,\text{and} \\
& \text{Kinetic}\,\text{Energy}\,\text{(}K.E.)=\dfrac{1}{2}\left| P.E. \right| \\
\end{align}$
Complete solution Step-by-Step:
The formula of potential energy is:
$\text{Potential Energy (}P.E.)=\dfrac{k{{q}^{2}}}{r}$, where $k$ is a constant value.
Now, we know that potential energy is negative for electrons and $q=e$, as given in the question. So, the potential energy of electron in hydrogen atom with charge $e$ and radius $r$ is:
$\text{Potential Energy (}P.E.)=\dfrac{-k{{e}^{2}}}{r}$
Since, potential energy of the electron in hydrogen atom is deduced above, now we can get the kinetic energy of electron in hydrogen atom using the formula mentioned below:
$\text{Kinetic}\,\text{Energy}\,\text{(}K.E.)=\dfrac{1}{2}\left| P.E. \right|$
Substituting the value of potential energy in the above formula, we get:
$\begin{align}
& K.E.=\dfrac{1}{2}\left| P.E. \right| \\
& \Rightarrow K.E.=\dfrac{1}{2}\left| \dfrac{-k{{e}^{2}}}{r} \right| \\
& \Rightarrow K.E.=\dfrac{1}{2}\times \dfrac{k{{e}^{2}}}{r} \\
& \therefore K.E.=\dfrac{k{{e}^{2}}}{2r} \\
\end{align}$
From the above deduced value,
$K.E.\propto \dfrac{{{e}^{2}}}{r}$
Therefore, the correct answer is Option (C).
Additional Information:
In physics, the kinetic energy (KE) of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body when decelerating from its current speed to a state of rest.
Note:
This is not the only way possible to solve this type of problems. There are different ways in which the answer can be achieved and feel free to use the way you are comfortable, but remember the formula for whichever way you choose, as if the formula is now known, then getting the correct answer becomes difficult.
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