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It is found experimentally that 13.6 eV energy is required to separate a hydrogen atom into a proton and an electron. Compute the orbital radius and the velocity of the electron in a hydrogen atom.

I know the total energy of an electron in a hydrogen atom but what does this statement mean?

It is found experimentally that 13.6 eV energy is required to separate a hydrogen atom into a proton and an electron

I am confused by this part.My textbook has equated this value with total energy of electron of a hydrogen atom but I didn't understand the reason

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The statement means just what it says - you need to input 13.6 eV to dismantle the atom into its component parts.

Any system of objects held together by attractive forces has associated with it what is known as a binding energy. This is the energy that is required to destroy the system, that is, to tear all its components apart and scatter them to infinite separation, and the energy that would be released by forming the system by allowing its components to attract each other under those interactive forces. The reason that energy is required to separate a system under the influence of attractive forces is that you have to supply force of your own to counteract those forces and move the components apart, and as you move objects over a distance with a force, you are doing work ($W = Fd$). Likewise, when the object is being formed, the attractive forces are themselves doing work, speeding up the constitutents. Quantum mechanics complicates the picture somewhat for the atomic case, but it is possible to formulate a suitable analogue of this.

In particular, if you started with a proton and electron at near-infinite distance, they would begin to accelerate toward each other and eventually would give off photons to end up with a total release of 13.6 eV and a hydrogen atom at the end. The giving off of photons is important - to settle into a bound system, the kinetic energy of components must be shed. Radiation is the mechanism by which this occurs.

A much larger example of the very same thing is a planet, such as the Earth. In this case the binding force is gravitational instead of electromagnetic, and one can talk of the gravitational binding energy. For the Earth, this is around $2.4 \times 10^{32}\ \mathrm{J}$. This is both the energy that would be required to blow the Earth apart as in the movie "Star Wars" and many similar ones, and it is also the energy that was released when the Earth first formed from the matter disk orbiting the primeval Sun about 4.54 billion years ago.Some of this energy persists on Earth today; that remainder is approximately one half of the current thermal reservoir within the planet, the other half being that produced subsequent to formation by the decay of radioactive materials, such as uranium contained within. Most of the energy was radiated into space as photons, just as with the atom, though through a much more complex process.

Regarding the relationship to the total energy, the total energy of a system is the sum of its kinetic and potential energy:

$$E_\mathrm{tot} = K + U$$

If the system is bound, then $E_\mathrm{tot}$ is less than zero, and if $E_\mathrm{tot}$ is zero or larger, the system is unbound. Thus the binding energy of the system will equal the energy you have to put in to cause $E_\mathrm{tot}$ to equal zero, and that is just its negative; and this is how 13.6 eV can be both the binding energy - the energy required to dismantle the system - and the negative of the system's total energy.

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  • $\begingroup$ So the whole system ie the nucleus and the electron are bounded by 13.6ev which is it's binding energy and the total energy of the system is equal to this $\endgroup$ Feb 19 '18 at 13:14
  • $\begingroup$ @The _Sympathizer Okay so the binding energy is the energy needed to form a system ..here the nucleus and electron and energy will be needed to destroy the system.But isn't the energy of the system fixed here so what's the point of equating that fixed enrgy with this binding energy $\endgroup$
    – user184272
    Feb 19 '18 at 14:06
  • $\begingroup$ @The_Sympathizer Nice answer but I didn't understand the last part $\endgroup$
    – user184271
    Feb 21 '18 at 10:09
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The hydrogen atom is a so-called "bound system". It takes energy to split it up into its constituents (in this case, an electron and a proton). Another example of a bound system is water. You can split H2O into H, H, and O, but it takes energy. Conversely, if you have H, H and O and assemble that into H2O, you release energy.

This is also why the 13.6 eV value is usually quoted with a minus sign. It is the energy that is released if you take a proton and an electron, infinitely far away from each other, and collapse them into a hydrogen atom. If you're familiar with gravitational potential energy, the concept is similar. Gravitational potential is negative, since energy is always released when two faraway objects fall towards each other.

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I think the question means to say that the work done by the nucleus' proton on the electron in bringing it from ∞ to that point is 13.6 eV. This will be the potential energy between the two charges:

$$ k\frac{q_1q_2}{r}= -13.6 \times 1.6 \times 10^{-19} J $$

k is coulomb's constant and can be approximated to 9*10^9.

1 eV = 1.6*10^-19 J

Both the charges will have the value of e.

For the velocity, take the centripetal force = coulomb's force (substituting the radius you found in the previous part)

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  • $\begingroup$ The first sentence is incorrect I'm afraid. The proton doesn't do work in keeping the electron in orbit. The -13.6 eV (note the negative sign) is the work done in taking an electron from infinity to the 1s orbital around the proton. $\endgroup$
    – Allure
    Feb 19 '18 at 8:59
  • $\begingroup$ Oh yes, you're right. But the potential energy formula will still apply right? $\endgroup$
    – Anshul
    Feb 19 '18 at 9:22
  • $\begingroup$ @Allure is it correct now? $\endgroup$
    – Anshul
    Feb 19 '18 at 9:29

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