EXAMPLE 28.1 The Balmer Series for Hydrogen Transitions responsible for the Balm

Question

EXAMPLE

28.1 The Balmer Series for Hydrogen

Transitions responsible for the Balmer series for the hydrogen atom. All transitions terminate at the n = 2 level.

GOAL Calculate the wavelength, frequency, and energy of a photon emitted during an electron transition in an atom.

PROBLEM The Balmer series for the hydrogen atom corresponds to electronic transitions that terminate in the state with quantum number n = 2, as shown in the figure (a) Find the longest-wavelength photon emitted in the Balmer series and determine its frequency and energy. (b) Find the shortest-wavelength photon emitted in the same series.

STRATEGY This problem is a matter of substituting values. The frequency can then be obtained from c= f and the energy from E = hf. The longest-wavelength photon corresponds to the one that is emitted when the electron jumps from the ni = 3 state to the nf = 2 state. The shortest-wavelength photon corresponds to the one that is emitted when the electron jumps from ni = to the nf = 2 state.

SOLUTION

(A) Find the longest-wavelength photon emitted in the Balmer series and determine its frequency and energy.

Substitute, with ni = 3 and nf = 2:

Take the reciprocal and substitute, finding the wavelength:

Now use c = f to obtain the frequency:

Calculate the photon's energy by substituting:

(B) Find the shortest-wavelength photon emitted in the Balmer series.

Substitute with 1/ni 0 as ni and nf = 2:

Take the reciprocal and substitute, finding the wavelength:

LEARN MORE

REMARKS The first wavelength is in the red region of the visible spectrum. We could also obtain the energy of the photon in the form hf = E3 - E2, where E2 and E3 are the energy levels of the hydrogen atom. Note that this photon is the lowest-energy photon in the Balmer series because it involves the smallest energy change. The second photon, the most energetic, is in the ultraviolet region.

QUESTION What is the upper-limit energy of a photon that can be emitted from hydrogen due to the transition of an electron between energy levels? (Select all that apply.)

1.70 eV13.6 eV(3/4) RH · h · cRH · h · c

PRACTICE IT

Use the worked example above to help you solve this problem. The Lyman series for the hydrogen atom corresponds to electronic transitions that terminate in the state with quantum number n = 1.(a) Find the longest-wavelength photon emitted in the Lyman series and determine its frequency and energy.

(b) Find the shortest-wavelength photon emitted in the same series.
nm

EXERCISEHINTS:  GETTING STARTED  |  I'M STUCK!

(a) Calculate the energy of the shortest wavelength photon emitted in the Balmer series for hydrogen.
E =  eV

(b) Calculate the wavelength of the photon emitted when an electron transits from n = 10 to n = 2.
=  nm

Explanation / Answer

We know that

Rydberg equation is 1/ =R(1/n2 -1/m2)

Where is the wavelength , R is Rydberg cosntant , m and n quantum number states

a)

For the longest wavelength photon emitted in the Lyman series and determine its frequency and energy is given by

n=1 and m=2

1/Lyman =1.097*107(1/12-1/22) =8227500m-1

Lyman =1.22*10-7m =122nm

The frequency is given by

v =f*lamda

f =c/lamda =3*108/1.22*10-7m =2.459*1015Hz

Energy is given by

E =hc/1.22*10-7m =6.625*10-34*3*108/1.22*10-7m =16.290*10-19/1.6*10-19 =10.18eV

For the shortest wavelenth is given by

n=1 and m=

1/Lyman = 1.097*107 * (1/12 - 1/2) = 10970000 m-1

Lyman = 9.12 x 10-8=91.2nm

a)

For the shortest wavelenth in Balmer series for hydrogen. is given by

n=2 and m=

1/Lyman = 1.097*107 * (1/22 - 1/2) = 2742500 m-1

Lyman = 3.6463 x 10-7=364.63nm

Therefore the energy is given by

E =hc/lamda =hc/3.6463 x 10-7m =6.625*10-34*3*108/3.6463*10-7m =5.4507*10-19/1.6*10-19 =3.4066eV

b)

Calculate the wavelength of the photon emitted when an electron transits from n = 10 to n = 2.
=  nm is

Rydberg equation is 1/ =R(1/n2 -1/m2)

1/ =1.097*107(1/22-1/102) =2632800m-1

=3.7982*10-7m =379.82nm

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