# Why is it that energy of electrons in higher orbits is greater than the ones in lower orbits?

There is this picture I found online which indicates that the energy level is higher as we go up and more electrons are added to the atom.

What I'm having trouble understanding is why the energy level is higher and not lower. The electrons are further and further away from the nucleus, at infinity their energy would be zero. Is it correct to say that the electrons' energy is negative and as we go further it increases, going to zero at infinity?

That is the only way I can see all this make sense.

• All energy values in the graph are negative, thus stabilisation. The zero energy of vacuum is somewhere over the top of it all. – Jan Nov 19 '17 at 15:09

What I'm having trouble understanding is why the energy level is higher and not lower. The electrons are further and further away from the nucleus, at infinity their energy would be zero.

You are correct. The innermost electrons are the most tightly bound to the nucleus and the outermost ('valence') electrons have the least energy, explaining why they can be removed (in chemical reactions e.g.)

But these energies are always negative and use the principle that infinitely far away from the nucleus the electron's energy would be $0$.

This way defined, the energy of the hydrogen electron in the ground state ($1s$) e.g. is $-13.6\ \mathrm{eV}$. The energy of the electron in excited states ($2s$, $2p$, $3s$ etc) is less negative, until for infinite distance (a free electron) it is zero ($0$).

Let's take a numerical example to demonstrate this. The ground state ($n=1$) of hydrogen's electron is $E_1=-13.6\ \mathrm{eV}$ and the first excited state ($n=2$) is $E_2=-3.4\ \mathrm{eV}$. What are the energies needed to remove the electron from these states?

Since as the energy at infinite distance is $0$, in the case $n=1$, then:

$$\Delta E= 0-(-13.6)=13.6\ \mathrm{eV}$$

And in the case $n=2$, then:

$$\Delta E= 0-(-3.4)=3.4\ \mathrm{eV}$$

So it takes less energy to remove an electron in a higher orbital than a lower one. That fits our observations.

Equally, the energy needed to excite the electron from $n=1$ to $n=2$ is:

$$\Delta E_{1\to 2}=-3.4-(-13.6)=10.2\ \mathrm{eV}$$

@Gert has given a great explanation and I like to give a small method for finding the answer for your question.

Consider an object which is 2Kg is on a level of 10m high from the considering level. It is drop from there and you know that it release its potential energy of mgh=200J. If it is on 20m high it release 400J. So we can see that

*when something drop from high energy level to a lower energy level it release its energy

*When it is drop from a more high energy level it release more energy than the other.

So, now we consider the Hydrogen spectrum of emission,

I will take some lines from that. Consider the line which is the reason for

3 →2
4 →2,

5 →2

According to the energy release 5 →2>4 →2>3 →2. So we can understand that 5th level is in a higher level 4 is then and 3 is then. Now as the above example the 5th level has a great energy than the 4th level.(Such like 20m level has more energy than the 10m level).

The nucleus attraction also affect for the value of the energy of the energy levels. As it goes above the attraction is coming lower due to that the gap of energy levels is coming lower.

As a summery,

When go above energy of the levels be increase, but the gap of the energy levels be decreased.

• The potential energy (given by mgh) should be ≈ 2kg*10m*10 ms^(-2) = 200 J, not 2kJ. – Tan Yong Boon Dec 19 '17 at 11:48
• Thank you for showing it. I will edit it. Sorry for have a such mistake. – Osal Thuduwage Dec 19 '17 at 13:03