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E n = 1 2 m e v 2 k Zq e 2 r n . size 12{E rSub { size 8{n} } = { {1} over {2} } m rSub { size 8{e} } v rSup { size 8{2} } - k { { ital "Zq" rSub { size 8{e} } rSup { size 8{2} } } over {r rSub { size 8{n} } } } } {}

Now we substitute r n size 12{r rSub { size 8{n} } } {} and v size 12{v} {} from earlier equations into the above expression for energy. Algebraic manipulation yields

E n = Z 2 n 2 E 0 ( n = 1, 2, 3, ... ) size 12{E rSub { size 8{n} } = - { {Z rSup { size 8{2} } } over {n rSup { size 8{2} } } } E rSub { size 8{0} } \( n=1," 2, 3, " "." "." "." \) } {}

for the orbital energies of hydrogen-like atoms    . Here, E 0 size 12{E rSub { size 8{0} } } {} is the ground-state energy n = 1 size 12{ left (n=1 right )} {} for hydrogen Z = 1 size 12{ left (Z=1 right )} {} and is given by

E 0 = 2 π 2 q e 4 m e k 2 h 2 = 13.6 eV.

Thus, for hydrogen,

E n = 13.6 eV n 2 ( n = 1, 2, 3, ...).

[link] shows an energy-level diagram for hydrogen that also illustrates how the various spectral series for hydrogen are related to transitions between energy levels.

An energy level diagram is shown. At the left, there is a vertical arrow showing the energy levels increasing from bottom to top. At the bottom, there is a horizontal line showing the energy levels of Lyman series, n is one. The energy is marked as negative thirteen point six electron volt. Then, in the upper half of the figure, another horizontal line showing Balmer series is shown when the value of n is two. The energy level is labeled as negative three point four zero electron volt. Above it there is another horizontal line showing Paschen series. The energy level is marked as negative one point five one electron volt. Above this line, some more lines are shown in a small area to show energy levels of other values of n.
Energy-level diagram for hydrogen showing the Lyman, Balmer, and Paschen series of transitions. The orbital energies are calculated using the above equation, first derived by Bohr.

Electron total energies are negative, since the electron is bound to the nucleus, analogous to being in a hole without enough kinetic energy to escape. As n size 12{n} {} approaches infinity, the total energy becomes zero. This corresponds to a free electron with no kinetic energy, since r n size 12{r rSub { size 8{n} } } {} gets very large for large n size 12{n} {} , and the electric potential energy thus becomes zero. Thus, 13.6 eV is needed to ionize hydrogen (to go from –13.6 eV to 0, or unbound), an experimentally verified number. Given more energy, the electron becomes unbound with some kinetic energy. For example, giving 15.0 eV to an electron in the ground state of hydrogen strips it from the atom and leaves it with 1.4 eV of kinetic energy.

Finally, let us consider the energy of a photon emitted in a downward transition, given by the equation to be

Δ E = hf = E i E f . size 12{ΔE= ital "hf"=E rSub { size 8{i} } - E rSub { size 8{f} } } {}

Substituting E n = ( 13.6 eV / n 2 ) size 12{E rSub { size 8{n} } = - "13" "." 6``"eV"/n rSup { size 8{2} } } {} , we see that

hf = 13.6 eV 1 n f 2 1 n i 2 . size 12{ ital "hf"= left ("13" "." 6" eV" right ) left ( { {1} over {n rSub { size 8{f} } rSup { size 8{2} } } } - { {1} over {n rSub { size 8{i} } rSup { size 8{2} } } } right )} {}

Dividing both sides of this equation by hc size 12{ ital "hc"} {} gives an expression for 1 / λ size 12{1/λ} {} :

hf hc = f c = 1 λ = 13.6 eV hc 1 n f 2 1 n i 2 . size 12{ { { ital "hf"} over { ital "hc"} } = { {f} over {c} } = { {1} over {λ} } = { { left ("13" "." 6" eV" right )} over { ital "hc"} } left ( { {1} over {n rSub { size 8{f} } rSup { size 8{2} } } } - { {1} over {n rSub { size 8{i} } rSup { size 8{2} } } } right )} {}

It can be shown that

13.6 eV hc = 13.6 eV 1.602 × 10 −19 J/eV 6.626 × 10 −34 J·s 2.998 × 10 8 m/s = 1.097 × 10 7 m –1 = R

is the Rydberg constant    . Thus, we have used Bohr’s assumptions to derive the formula first proposed by Balmer years earlier as a recipe to fit experimental data.

1 λ = R 1 n f 2 1 n i 2 size 12{ { {1} over {λ} } =R left ( { {1} over {n rSub { size 8{f} } rSup { size 8{2} } } } - { {1} over {n rSub { size 8{i} } rSup { size 8{2} } } } right )} {}

We see that Bohr’s theory of the hydrogen atom answers the question as to why this previously known formula describes the hydrogen spectrum. It is because the energy levels are proportional to 1 / n 2 size 12{1/n rSup { size 8{2} } } {} , where n size 12{n} {} is a non-negative integer. A downward transition releases energy, and so n i size 12{n rSub { size 8{i} } } {} must be greater than n f size 12{n rSub { size 8{f} } } {} . The various series are those where the transitions end on a certain level. For the Lyman series, n f = 1 size 12{n rSub { size 8{f} } =1} {} — that is, all the transitions end in the ground state (see also [link] ). For the Balmer series, n f = 2 size 12{n rSub { size 8{f} } =2} {} , or all the transitions end in the first excited state; and so on. What was once a recipe is now based in physics, and something new is emerging—angular momentum is quantized.

Triumphs and limits of the bohr theory

Bohr did what no one had been able to do before. Not only did he explain the spectrum of hydrogen, he correctly calculated the size of the atom from basic physics. Some of his ideas are broadly applicable. Electron orbital energies are quantized in all atoms and molecules. Angular momentum is quantized. The electrons do not spiral into the nucleus, as expected classically (accelerated charges radiate, so that the electron orbits classically would decay quickly, and the electrons would sit on the nucleus—matter would collapse). These are major triumphs.

Questions & Answers

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smart Reply
what is velocity
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displacement per unit time
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the ratec of displacement over time
Jamie
the rate of displacement over time
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the rate of displacement over time
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up to tomorrow
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i need a description and derivation of kinetic theory of gas
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did you need it right now
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Faith Reply
Aunt Faith,please i am thinking the dissolution here from the word "solution" exposed the grains of salt to be dissolved in the water.Thankyou
Junior
dissolution please
Junior
Aunt Faith,please i am thinking the dissolution here from the word "solution" exposed the grains of salt to be dissolved in the water.Thankyou
Junior
it is either diffusion or osmosis. just confused
Faith
due to solvation....
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what is solvation pls
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water molecule surround the salt molecules . solute solute attraction break in the same manner solvent solvent interaction also break. as a result solute and solvent attraction took place.
Pathani
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Faith
my pleasure
Pathani
what is solvation pls
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Faith
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Pathani
due to solvation....
Pathani
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physical phenomena arising from force caused by magnets
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is the phenomenon of attracting magnetic substance like iron, cobalt etc.
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John Reply
Heat is a form of energy where molecules move
saran
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topic-- question
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Lilian Reply
20cm3 of 1mol/dm3 solution of a monobasic acid HA and 20cm3 of 1mol/dm3 solution of NaOH are mixed in a calorimeter and a temperature rise of 274K is observed. If the heat capacity of the calorimeter is 160J/K, calculate the enthalpy of neutralization of the acid.(SHCw=4.2J/g/K) Formula. (ms*cs+C)*T
Lilian
because it changes only direction and the speed is kept constant
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Practice Key Terms 7

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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