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I am currently in the 11th grade. In the Bohr Model of the hydrogen atom, potential energy is two times that of total energy. It means that magnitude of potential energy is two times that of magnitude of total energy(or not?). But how is this possible as we are considering potential energy as a part of total energy? (I know the mathematical derivation behind it and I also know that the potential energy is negative in this case, but still it doesn't make any sense to me to have potential energy twice the total energy.) Sorry, if the question is silly.

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In the Bohr model of the hydrogen atom we take the potential energy of the electron to be zero at infinity, so the potential energy becomes negative as the electron approaches the hydrogen atom. However kinetic energy is always positive.

In the Bohr ground state the potential energy is -27.2 eV. Note that as described above this energy is negative. The kinetic energy is +13.6eV, so when we add the two together we get the total energy to be -13.6eV.

The total energy is negative because the electron is bound to the hydrogen atom and to remove the electron we have to put in energy. In fact we have to put in 13.6eV, which is simply the ionisation energy of hydrogen. Since the ground state has an energy of -13.6eV when we put in +13.6eV that makes the total energy zero, and as we said at the start this is the energy of the electron at infinity.

In any system where the force obeys an inverse square law there is a link between the potential and kinetic energy. This link is called the virial theorem and it says that if we call the potential energy $V$ and the kinetic energy $T$ the relationship is:

$$ V = -2T $$

Note the minus sign in this equation. The total energy $E$ is $V+T$, and if we add $T$ to both sides of this equation we get:

$$ E = V + T = -T = V/2 $$

That is why the total energy is half the potential energy.

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  • $\begingroup$ But average potential and average kinetic energy are related to system averages. So how electrons kinetic energy be related to systems kinetic energy? $\endgroup$ Commented Feb 13, 2020 at 18:24

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