The Rydberg’s equation for the hydrogen atom is:

1 = 1.097 � 107

( 1 – 1 )λ n12 n2

2

Use the above equation to calculate the (i) wave number, (ii) wavelength and (iii) frequency forthe last line in the Balmer series.

Solution 3.1Using n1 = 2 and n2 = ∞

1 = 1.097 � 107

( 1 – 1 )λ 22 ∞2

(i)1

= 2.74 � 106 m–1

λ(ii) λ = 3.65 � 10–7 m or = 365 nm (Wavelength)

(iii) f = Cλ

= 3.0 � 108

3.65 � 10–7

= 8.22 � 1014 Hz (frequency)

In the stratosphere, chlorine molecules absorb electromagnetic radiation and dissociate to produce chlorine atoms, which are responsible for the destruction of ozone.

Cl—Cl(g) → 2Cl(g)

Given that the bond energy of chlorine is 242 kJ mol–1, calculate the wavelength of the lightabsorbed.

Solution 3.2E = hf

242 = (3.99 � 10–13) � ff = 6.07 � 1014 Hz

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CHAPTER 3 ATOMIC STRUCTURE

Question 3.1

Question 3.2

Wavelength = Cf

= 3.0 � 108

6.07 � 1014

= 4.94 � 10–7 mOr = 494 nm

Part of the Lyman series in the hydrogen emission spectrum is as shown below:

(a) Draw an energy level diagram to show how the lines are produced.(b) (i) Define ionisation energy of hydrogen. Illustrate your answer with an appropriate

equation.(ii) Calculate the ionisation energy (in kJ mol–1) of hydrogen from the above spectrum.

Solution 3.3(a)

(b) (i) Ionisation energy of hydrogen is the minimum energy required to remove the loneelectron in the hydrogen atom per mole of gaseous hydrogen atom under standard conditions (298 K and 101 kPa).

H(g) → H+(g) + e

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Question 3.3

88.2 10.27 10.62 10.97

Wave number ( x 106 ) m–1

88.2 10.27 10.62 10.97

Wave number ( x 106 ) m–1

n = 1

n = 2

n = 3

n = 4

n = �

(ii) E = hf

= h Cλ

= (3.99 � 10–13)(3.0 � 108)(10.97 � 106)= 1313.1 kJ mol–1

The diagram below shows the lines in the Balmer series in the emission spectrum of hydrogenatom.

(a) Draw a labelled energy diagram to show how the lines marked X and Y are formed.(b) Name two species that would produce similar emission spectrum as that of the hydrogen

atom. Explain your reasoning.

Solution 3.4(a)

(b) Li2+ and He+. Both the ions are one electron species.

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Question 3.4

X Y

Frequency

X Y

Frequency

n = 1

n = 2

n = 3

n = 4

n = �

(a) Explain the difference between orbit and orbital.(b) Draw the shape of

(i) an s-orbital(ii) a p-orbital

(iii) the five d-orbitals

Solution 3.5(a) Orbital refers to the volume in space around the nucleus where the probability of finding a

particular electron in � 95%.Orbit refers to the fixed circular path around the nucleus where a particular electronrevolves.

(b) (i) s-orbital

(ii) p-orbitals

(iii) d-orbitals

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Question 3.5

y

z

x

Y

X

Z

z z z z z

y y y y y

x x xx x

dxz dye dxy

dx2-y2

dx2