
Suppose an electron is attracted towards the origin by a force \[\dfrac{k}{r}\] where 'k' is a constant and 'r' is the distance of the electron from the origin. By applying Bohr model to this system, the radius of the nth orbital of the electron is found to be ′\[{r_n}\]′ and the kinetic energy of the electron to be ′\[{T_n}\] ′. Then which of the following is true?
A. \[{T_n}\] independent of n, \[{r_n} \propto n\]
B. \[{T_n} \propto \dfrac{1}{n},{r_n} \propto n\]
C. \[{T_n} \propto \dfrac{1}{n}{n_1},{r_n} \propto {n^2}\]
D. \[{T_n} \propto \dfrac{1}{{{n^2}}},{r_n} \propto {n^2}\]
Answer
163.8k+ views
Hint:Using Bohr’s model of atomic structure, the electron revolves around the nucleus of the atom in a fixed orbit having fixed energy. As the orbit is fixed so the distance of the orbit is fixed from the center of the atom. Hence, the electron is in equilibrium along the radial direction.
Formula used:
\[{F_o} = \dfrac{{m{v^2}}}{r}\]
where \[{F_o}\] is the outward centrifugal force acting on a body of mass m revolving in a circular orbit of radius r with speed v.
\[T = \dfrac{{m{v^2}}}{2}\]
where T is the kinetic energy of the body of mass m with speed v.
Complete step by step solution:
The given centripetal force is \[\dfrac{k}{r}\] where r is the radius of the orbit and k is any constant. To move in a circular orbit of fixed radius, the net force acting on the electron must be zero along the radial direction. So, the magnitude of the centripetal force must be equal to the magnitude of the centrifugal force.
\[\dfrac{{m{v^2}}}{r} = \dfrac{k}{r} \\ \]
\[\Rightarrow v = \sqrt {\dfrac{k}{m}} \]
So, the speed of the electron in the orbit is \[\sqrt {\dfrac{k}{m}} \].
According to the Bohr’s model the angular momentum of the electron is quantized and it is given as,
\[L = \dfrac{{nh}}{{2\pi }}\]
The angular momentum is given as \[L = mvr\].
So, using both the expression,
\[mvr = \dfrac{{nh}}{{2\pi }} \\ \]
\[\Rightarrow r = \dfrac{{nh}}{{2\pi mv}} \\ \]
\[\Rightarrow r = \left( {\dfrac{h}{{2\pi m}}\sqrt {\dfrac{m}{k}} } \right)n\]
So, the radius of the nth energy level orbit is proportional to the principal quantum number n.
\[r \propto n\]
The kinetic energy of the electron is,
\[T = \dfrac{{m{v^2}}}{2} \\ \]
\[\Rightarrow T = \dfrac{{mk}}{{2m}} \\ \]
\[\therefore T = \dfrac{k}{2} = \]constant
Hence, the kinetic energy of the electron is independent of n.
Therefore, the correct option is A.
Note: If the energy of the orbit is not fixed then the electron in the orbit must be accelerating. And accelerating charge particles liberate energy and finally it will fall into the nucleus. This will destabilize the nucleus of the atom.
Formula used:
\[{F_o} = \dfrac{{m{v^2}}}{r}\]
where \[{F_o}\] is the outward centrifugal force acting on a body of mass m revolving in a circular orbit of radius r with speed v.
\[T = \dfrac{{m{v^2}}}{2}\]
where T is the kinetic energy of the body of mass m with speed v.
Complete step by step solution:
The given centripetal force is \[\dfrac{k}{r}\] where r is the radius of the orbit and k is any constant. To move in a circular orbit of fixed radius, the net force acting on the electron must be zero along the radial direction. So, the magnitude of the centripetal force must be equal to the magnitude of the centrifugal force.
\[\dfrac{{m{v^2}}}{r} = \dfrac{k}{r} \\ \]
\[\Rightarrow v = \sqrt {\dfrac{k}{m}} \]
So, the speed of the electron in the orbit is \[\sqrt {\dfrac{k}{m}} \].
According to the Bohr’s model the angular momentum of the electron is quantized and it is given as,
\[L = \dfrac{{nh}}{{2\pi }}\]
The angular momentum is given as \[L = mvr\].
So, using both the expression,
\[mvr = \dfrac{{nh}}{{2\pi }} \\ \]
\[\Rightarrow r = \dfrac{{nh}}{{2\pi mv}} \\ \]
\[\Rightarrow r = \left( {\dfrac{h}{{2\pi m}}\sqrt {\dfrac{m}{k}} } \right)n\]
So, the radius of the nth energy level orbit is proportional to the principal quantum number n.
\[r \propto n\]
The kinetic energy of the electron is,
\[T = \dfrac{{m{v^2}}}{2} \\ \]
\[\Rightarrow T = \dfrac{{mk}}{{2m}} \\ \]
\[\therefore T = \dfrac{k}{2} = \]constant
Hence, the kinetic energy of the electron is independent of n.
Therefore, the correct option is A.
Note: If the energy of the orbit is not fixed then the electron in the orbit must be accelerating. And accelerating charge particles liberate energy and finally it will fall into the nucleus. This will destabilize the nucleus of the atom.
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