
If the Bohr model of hydrogen atom, the electron is treated as a particle going in a circle with the centre at the proton. The proton itself is assumed to be fixed in an inertial frame. The centripetal force is provided by the Coulomb attraction. In the ground state, the electron goes round the proton in a circle of radius \[5.3\times {{10}^{-11}}m\]. find the speed of the electron in the ground state. Mass of the electron= \[9.1\times {{10}^{-31}}kg\] and charge of the electron= \[1.6\times {{10}^{-19}}C\]
Answer
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Hint: Centripetal force is defined as the component of force which is acted on the object in curvilinear motion and is directed towards the axis of rotation. The SI unit of centripetal force is Newton. It directed perpendicularly to the displacement of objects. The formula of centripetal force is the product of centripetal acceleration and mass of object.
Complete solution:
The electron is rotated around the proton in the circular direction with the proton at the centre. As it is given in question that the force is provided by coulomb attraction.
Radius= \[5.3\times {{10}^{-11}}m\]
mass of electron= \[9.1\times {{10}^{-31}}kg\]
charge of electron = \[1.6\times {{10}^{-19}}C\]
when the change in the particle is particular then the centripetal force is balanced by electrostatic force.
So, centripetal force= electrostatic force
The formula for both the forces are
\[\begin{align}
& \dfrac{m{{v}^{2}}}{r}=k\dfrac{{{q}^{2}}}{{{r}^{2}}} \\
& \Rightarrow {{v}^{2}}=\dfrac{k{{q}^{2}}}{rm} \\
\end{align}\]
Substituting the values in the above formula we get,
\[{{v}^{2}}=\dfrac{9\times {{10}^{9}}\times 1.6\times 1.6\times {{10}^{-38}}}{5.3\times {{10}^{-11}}\times 9.1\times {{10}^{-31}}}=\dfrac{23.04}{48.23}\times {{10}^{13}}\]
On simplify we get
\[{{v}^{2}}=4.077\times {{10}^{13}}=4.7\times {{10}^{12}}\]
On under rooting both sides we get
\[v=2.2\times {{10}^{6}}m/\sec \]
The speed of the electron is \[v=2.2\times {{10}^{6}}m/\sec \].
Note:the concept of centripetal force is used in spinning of the ball, turning a car, planets orbiting the Sun. the opposite of centripetal force is centrifugal force. it arises from the inertia of the body and appears as if it is acting on the body. It is used in making a bike turn or when the vehicle is turned around a curve.
Complete solution:
The electron is rotated around the proton in the circular direction with the proton at the centre. As it is given in question that the force is provided by coulomb attraction.
Radius= \[5.3\times {{10}^{-11}}m\]
mass of electron= \[9.1\times {{10}^{-31}}kg\]
charge of electron = \[1.6\times {{10}^{-19}}C\]
when the change in the particle is particular then the centripetal force is balanced by electrostatic force.
So, centripetal force= electrostatic force
The formula for both the forces are
\[\begin{align}
& \dfrac{m{{v}^{2}}}{r}=k\dfrac{{{q}^{2}}}{{{r}^{2}}} \\
& \Rightarrow {{v}^{2}}=\dfrac{k{{q}^{2}}}{rm} \\
\end{align}\]
Substituting the values in the above formula we get,
\[{{v}^{2}}=\dfrac{9\times {{10}^{9}}\times 1.6\times 1.6\times {{10}^{-38}}}{5.3\times {{10}^{-11}}\times 9.1\times {{10}^{-31}}}=\dfrac{23.04}{48.23}\times {{10}^{13}}\]
On simplify we get
\[{{v}^{2}}=4.077\times {{10}^{13}}=4.7\times {{10}^{12}}\]
On under rooting both sides we get
\[v=2.2\times {{10}^{6}}m/\sec \]
The speed of the electron is \[v=2.2\times {{10}^{6}}m/\sec \].
Note:the concept of centripetal force is used in spinning of the ball, turning a car, planets orbiting the Sun. the opposite of centripetal force is centrifugal force. it arises from the inertia of the body and appears as if it is acting on the body. It is used in making a bike turn or when the vehicle is turned around a curve.
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