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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

what is physics
Rhema Reply
a15kg powerexerted by the foresafter 3second
Firdos Reply
what is displacement
Xolani Reply
movement in a direction
Jason
hello
Hosea
Explain why magnetic damping might not be effective on an object made of several thin conducting layers separated by insulation? can someone please explain this i need it for my final exam
anas Reply
Hi
saeid
hi
Yimam
What is thê principle behind movement of thê taps control
Oluwakayode Reply
while
Hosea
what is atomic mass
thomas Reply
this is the mass of an atom of an element in ratio with the mass of carbon-atom
Chukwuka
show me how to get the accuracies of the values of the resistors for the two circuits i.e for series and parallel sides
Jesuovie Reply
Explain why it is difficult to have an ideal machine in real life situations.
Isaac Reply
tell me
Promise
what's the s . i unit for couple?
Promise
its s.i unit is Nm
Covenant
Force×perpendicular distance N×m=Nm
Oluwakayode
İt iş diffucult to have idêal machine because of FRİCTİON definitely reduce thê efficiency
Oluwakayode
if the classica theory of specific heat is valid,what would be the thermal energy of one kmol of copper at the debye temperature (for copper is 340k)
Zaharadeen Reply
can i get all formulas of physics
BPH Reply
yes
haider
what affects fluid
Doreen Reply
pressure
Oluwakayode
Dimension for force MLT-2
Promise Reply
what is the dimensions of Force?
Osueke Reply
how do you calculate the 5% uncertainty of 4cm?
melia Reply
4cm/100×5= 0.2cm
haider
how do you calculate the 5% absolute uncertainty of a 200g mass?
melia Reply
= 200g±(5%)10g
haider
use the 10g as the uncertainty?
melia
which topic u discussing about?
haider
topic of question?
haider
the relationship between the applied force and the deflection
melia
sorry wrong question i meant the 5% uncertainty of 4cm?
melia
its 0.2 cm or 2mm
haider
thank you
melia
Hello group...
Chioma
hi
haider
well hello there
sean
hi
Noks
hii
Chibueze
10g
Olokuntoye
0.2m
Olokuntoye
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thomas
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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