Although energy states differ in every type of atom, the farther an orbit from the nucleus, the energy. In normal situation, electrons will circle the nucleus at "ground state", the lowest energy state in which they can be present. However, when any type of energy, such as heat or light, is absorbed by the atom, these electrons will often "jump" to higher energy states, or "levels". Soon after, an electron will most likely jump back to ground state, and the atom will emit the same amount of energy it absorbed in the form of heat or light. The amount of energy it takes for an electron to jump energy levels is known as transition energy. The energy absorbed must exactly equal the transition energy for an electron to jump energy levels. So exactly how muc enegy is involved in a transition, and how does this relate to the distance of the energy level from the nucleus? The solution lies in the specific properties of the atom.
Electrons behave like all matter in their duality of wave and particle properties. Their properties of light are what make us understand the specific amounts of energy it takes for an electron transition to occur. When electrons orbit the nucleus, Bohr used wave mechanics to postulate that they travel in waves, particularly standing waves with definite nodes and antinodes. The reason why only specific states exist in an atom is because these are the only states when an electron can be described as a standing wave, with a specific number of nodes and antinodes. This was proposed as the "semi-classical" picture of the atom, where electrons traveled in standing waves at certain distances from the nucleus. In the semiclassical picture, these stationary states have electrons moving at a constant speed with a constant wavelength.
Both kinetic and potential energy must be examined when determining the transition energy of an electron. Before determining the energies, several variables and constants must be introduced:
m = mass (mass of electron = 9.1 * 10-31 kg) v = velocity
r = radius (distance between energy state and nucleus)
Z= nuclear charge, or amount of protons (Z=l for hydrogen with 1 proton)
TE = Total Energy
KE = Kinetic Energy
n = number of antinodes, and number of energy level
PE = potential energy
= wavelength
Q = charge of body
k = Coloumb's constant (k = 8.9 * 109 Nm2/C2)
e = charge of electron (1. 6 * 10-19 c)
h = Plank's constant, (h = 6.63 * 10-34 jS); H = "h bar", derivative of Plank's constant h / 2 
When an electron orbits the nucleus, it has a specific centripetal force which keeps it circling the nucleus. Newton's second law states that f orce is equal to the mass times the acceleration of an object. In this specific case, an electric force will provide the energy for the centripetal force. The centripetal force of an object is defined by this equation:
eq. 1 Fc = mv2 /r
and the electric force is defined as:
eq. 2 Fe = k (Q1) (Q2) / r2
So therefore:
Fc = Fe
eq. 3 mv2/ r = k (Q1) (Q2) / r2
Because the variable Z is equal to the number of protons in the nucleus, the charge of the nucleus can be expressed as Ze. If Ze and e, the charge of the electron, are plugged in as the two charges, the equation looks like this:
eq. 4 mv2/r = kZe2/r2
If the formula for kinetic energy (KE = 1/2MV2) is plugged into the equation, we find the kinetic energy of the system to be:
eq. 5 KE = kZe2/2ro
Ro is called "radius naught". It represents the radius at its initial value.
The potential energy of the system requires the use of integral calculus to derive an exact formula. Potential energy is equal to the change in work of a system. Work is equal to the force times the distance moved of a particular object. (W = Fd) In this case, we can not def ine r at a specif ic point, because it constantly changes its distance from the nucleus. Instead, using calculus, r must be calculated by computing the sum of each small distance that r moves. In this case, the variable d is equal to the change in r. So therefore:
eq. 6 
w = Fr (change in r)
Now we must take the integral of both sides, and substituting potential energy back in for work, the formula becomes:
eq. 7 PE = Fdr
(dr represents the change in r)
This is the integral of work between the values of r naught and infinity, as r gets infinitely close to 0.
Substituting the Force we found earlier for F, (F = kZe 2 /r 2 ) , and pulling out the constants kZe , the formula becomes:
eq. 8 PE = kZe2 dr/r2
Now we can take the integral with respect to the radius, finding the integral to be:
eq. 9 PE = kZe2(-1/r)
Evaluating it at both intervals, r naught and infinity, the potential energy of the system is:
eq. 10 PE = -kZe2 / ro
Once the kinetic and potential energy are added together, the total energy of the system is:
eq. 1 1 TE = -kZe2 / 2ro
The total energy of the system is negative because this atom is what is known as a "bound system", where energies inside the system must always be negative. The negative sign comes from the potential energy formula because the change in work is actually negative; the direction of the work when an electron jumps directly opposes the force that pulls it towards the nucleus.
Bohr wanted to find constant radii that would support his theory of stationary waves and states of electrons. He came up with a third postulate that would solve this problem. He postulated that the angular momentum of an electron in orbit equaled Plank's constant times n, the number of antinodes in the standing wave, or the energy level number. The angular momentum of any circular orbit is equal to mvr, so therefore:
eq. 12 mvr = nh/2p
Bohr used his stationary wavelengths to postulate this formula. The standing waves of the electrons in orbit all have a specific wavelength. Another physicist of the time, L. de Broglie, found that the wavelength of any object is equal to Planck's constant over the momentum, of the wave:
eq. 13 = h/p
The momentum of any particle is equal to mass times velocity so substituting for p, the equation becomes:
eq. 14 = h/mv
If this wave traveled in a complete circular orbit, as in the "semi-classical" picture of the energy states, the wavelength would also have a specific circumference directly related to the whole number of antinodes. Now, the equation is,
eq. 15 n = 2r
If we now substitute (h/mv) for the wavelength here, and rearrange the formula slightly, Bohr's third postulate proves true given that electrons exist in standing wave patterns:
eq. 16 mvr = nh / 2
To simplify variables, "h bar" is a constant equivalent to Planck's constant divided by 2 so,
eq. 17 mvr = n
If we solve for v, the velocity, we find that
eq. 18 v = n / rm
Now, if v is plugged back into the kinetic energy formula (KE = 1/2 mv2) , and the new kinetic energy is used, the formula becomes,
eq. 19 KE = m/2 (n/rm ) 2= kZe2/ 2r
If the radius (r) is isolated in this equation,
eq. 2 0 r = n22 / kZe2
To simplify the excessive amount of constants, Bohr came up with a single constant that would represent them all, "A naught". A naught is equal to: Ao = l2/ kZe2 and therefore the radius is equal to n2Ao. Ao is known as the Bohr radius, the smallest radius in a Hydrogen atom from the nucleus to first ground state. Its numerical value is .0529 nm.
After finding the values of specific radii, the last step is to find the value of each actual energy level in the semi-classical hydrogen atom. Equation #11 dictates that the energy of the system is
TE = -kZe2 / 2ro. Using the Bohr radius, it is possible to f ind levels of particular stationary states.
Plug r = n2Ao into the total energy equation. The equation now becomes,
eq. 21 E = -kZe2/ 2n2Ao
Bohr then defined another constant to simplify the number of constants. He called the set of constants kZe2 /2Ao, E naught (Eo) . Eo is equal to about 13.6 Ev's. An Ev is a unit for energy that is equal to 1.6 * 10-19 joules, So then the final equation for the specific energy levels in a hydrogen atom is:
E = -Eo / n2
The first three energy levels of Hydrogen, are -13.69 Ev's (ground state) , -3.43 Ev's, and -1.52 Ev's, respectively. Although the energy levels appear to decrease, they are actually increasing as they approach zero. While the energy is increasing, the rate of change is also increasing. There is a greater difference in energy between ground state and the second energy level than that of the second and third energy levels. Electrons tend to exist in the ground state because of the relatively large amount of energy binding them.
All these atomic energies are in theory related to the atomic energies of the dye used in the Tunable Dye laser. When the dye absorbs energy becomes excited, its specific molecules undergo energy transitions which are much more complicated than that of the simple hydrogen atom. The dye being used, Rhodamine 6G, oscillates after absorbing energy. This flex energy (E = 1/2kx2) is similar to the kinetic energy formula utilized by Bohr, where x is the level of flexing or oscillation. Transition energy is important to help understand the excitation of the dye and the amplification of light on a molecular level.