PROBLEM 111P

You learned in this chapter that ionization generally increases as you move from left to right across the periodic table. However, consider the data below, which shows the ionization energies of the period 2 and 3 elements:

Group |
Period 2 Elements |
Ionization Energy (kJ/mol) |
Period 3 Elements |
Ionization Energy (kJ/mol) |

1A |
Li |
520 |
Na |
496 |

2A |
Be |
899 |
Mg |
738 |

3A |
B |
801 |
Al |
578 |

4A |
C |
1086 |
Si |
786 |

5A |
N |
1402 |
P |
1012 |

6A |
O |
1314 |
S |
1000 |

7A |
F |
1681 |
Cl |
1251 |

8A |
Ne |
2081 |
Ar |
1521 |

Notice that the increase is not uniform. In fact, ionization energy actually decreases a bit in going from elements in group 2A to 3A and then again from 5A to 6A. Use what you know about electron configurations to explain why these dips in ionization energy exist.

Solution 110P

Here, we have to calculate the energy of 1 mol of photons with a wavelength of 632 nm.

We know that

A photon of wavelength λ has an energy (E) given by the equation:

E = hc/λ, -----(1)

where E is the energy of the photon in J, h is Planck's constant and c is the speed of light.

Given that,

Planck's constant h = 6.626 × 10−34 J• s

Speed of Light c = (3.00 × 108 m/s)

Wavelength of light = 632 nm = 632 x 10-9 m

Thus, from equation (1)

E = (6.626 × 10−34 J• s x 3.00 × 108 m/s)/ 632 x 10-9 m

= 0.0314 x 10−17 J

Therefore,

1 mol photon = 6.022 x 1023 photon

Thus, the energy of 1 mol of photons = (6.022 x 1023 x 0.0314 x 10−17 J)

= 0.1890 x 106 J = 189 kJ

Results: The energy of 1 mol of photons with a wavelength of 632 nm is 189 kJ