
If m and e are the mass and charge of the revolving electron in the orbit of radius r for hydrogen atom, the total energy of the revolving electron will be:
\[
{\text{A}}{\text{. }}\dfrac{1}{2}\dfrac{{k{e^2}}}{r} \\
{\text{B}}{\text{. }} - \dfrac{{k{e^2}}}{r} \\
{\text{C}}{\text{. }}\dfrac{{mk{e^2}}}{r} \\
{\text{D}}{\text{. }} - \dfrac{1}{2}\dfrac{{k{e^2}}}{r} \\
\]
Answer
567k+ views
Hint: The total energy of the revolving electron will be equal to the sum of the kinetic energy and the potential energy of the electron. The kinetic energy is provided by the centripetal force which is provided by the force of attraction between the nucleus and the electron. The potential energy is equal to the work done to bring electrons from infinity to the given orbit around the nucleus.
Complete answer:
We are given an electron of mass m and charge e revolving around the nucleus in an orbit of radius r in a hydrogen atom.
The kinetic energy of the electron revolving around the nucleus is due to the centripetal force acting on the electron. This centripetal force is provided by the Coulombic attraction between the negatively charged electron and the positively charged nucleus.
Therefore, the kinetic energy of an electron revolving around nucleus is given as
$K.E. = \dfrac{1}{2}\dfrac{{Zk{e^2}}}{r}$
Here Z represents the atomic number which is equal to one for hydrogen atom and k is given as $k = \dfrac{1}{{4\pi { \in _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}}$.
Now the potential energy of an electron revolving around the nucleus is equal to the work done to bring an electron from infinity to the given orbit around the nucleus. It is given as
$P.E. = - \dfrac{{Zk{e^2}}}{r}$
Now the total energy of the electron revolving around the nucleus is equal to the sum of the kinetic energy and the potential energy of the electron. Therefore, we have
$
T.E. = K.E + P.E \\
= \dfrac{1}{2}\dfrac{{k{e^2}}}{r} + \left( { - \dfrac{{k{e^2}}}{r}} \right) \\
= - \dfrac{1}{2}\dfrac{{k{e^2}}}{r} \\
$
This is the required answer.
Therefore, the correct answer is option D.
Note:
These relations can also describe the energy of the electron in an hydrogen like atom. These are atoms which have electronic configuration similar to that of the hydrogen atom like He$^ + $, Li$^{ + + }$, etc. In that case we need to change the value of atomic number Z depending on the atom under consideration.
Complete answer:
We are given an electron of mass m and charge e revolving around the nucleus in an orbit of radius r in a hydrogen atom.
The kinetic energy of the electron revolving around the nucleus is due to the centripetal force acting on the electron. This centripetal force is provided by the Coulombic attraction between the negatively charged electron and the positively charged nucleus.
Therefore, the kinetic energy of an electron revolving around nucleus is given as
$K.E. = \dfrac{1}{2}\dfrac{{Zk{e^2}}}{r}$
Here Z represents the atomic number which is equal to one for hydrogen atom and k is given as $k = \dfrac{1}{{4\pi { \in _0}}} = 9 \times {10^9}N{m^2}{C^{ - 2}}$.
Now the potential energy of an electron revolving around the nucleus is equal to the work done to bring an electron from infinity to the given orbit around the nucleus. It is given as
$P.E. = - \dfrac{{Zk{e^2}}}{r}$
Now the total energy of the electron revolving around the nucleus is equal to the sum of the kinetic energy and the potential energy of the electron. Therefore, we have
$
T.E. = K.E + P.E \\
= \dfrac{1}{2}\dfrac{{k{e^2}}}{r} + \left( { - \dfrac{{k{e^2}}}{r}} \right) \\
= - \dfrac{1}{2}\dfrac{{k{e^2}}}{r} \\
$
This is the required answer.
Therefore, the correct answer is option D.
Note:
These relations can also describe the energy of the electron in an hydrogen like atom. These are atoms which have electronic configuration similar to that of the hydrogen atom like He$^ + $, Li$^{ + + }$, etc. In that case we need to change the value of atomic number Z depending on the atom under consideration.
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