Answers to Physics Bridging Course
Chapter 8
1.
The radius of a gold nucleus would be of the order of 10-15 metres.
2.
Although the volume of any atom is almost entirely empty space, it does possess structure.
Forces, mainly electrostatic forces, penetrate the vacuum. Attractive forces between the nucleus
and each electron hold the atom together, while repulsive forces between electrons help to
prevent an atom from collapsing, and also prevent other atoms from passing through.
3.
Such diagrams are correct in that there is a nucleus containing protons and neutrons in the
centre of the atom, with electrons in shells surrounding it having discrete energy levels.
However, the diagrams are not to scale, atoms are 3-D structures close to spherical, certainly
not 2-D circles, and electrons are no longer believed to orbit the nucleus in circular orbits.
4.
E = hf
 Eradio < 6.63  10-34  109 = 6.63  10-25 J  1.6  10-19 < 4.1  10-6 eV
Similarly, Eμw < 1.2  10-3 eV
EX-rays < 0.12 MeV
5.
(a) E  h f 

EI-R < 1.8 eV
Evis < 3.1 eV
EU-V < 600 eV
Eγ > 0.12 MeV (but there is some cross-over).
 
hc
6.63 10 34  2.98  10 8

λ
532  10 9



 3.714  10 19 J
(b) 1 W = 1 J s
 n  3.714  10 = 1
 n = 2.7  1018 photons/second
Since red light has a longer wavelength than green light, its frequency is less, so the energy
of its photons must be less as well. Hence, a far greater number of red photons are needed.
-1
-19
6.
The energy of the photons of U-V light is sufficient to damage skin cells, and cause mutations
in the DNA which often causes cancer. The photons of I-R radiation are unable to do this since
their energy is too low, and although there may be more I-R photons, they are still harmless.
7.
Electrical potential energy is lost by an electron when it is “trapped” by a hydrogen ion, and
drops into the ground state. Since it began with U = 0, it must now have negative energy.
Considerable energy would need to be provided for that electron to escape from its atom.
8.
Using the Rydberg formula
= 8.2275  107
 1
1
1 
1
1 
1
 RH  2  2  
 1.097  10 7  2  2 
n
λ
λ1
2 
ni 
1
 f
 λ1 = 1.21  10-7 m
 E1 = – 10.2 eV
Similarly, λ2 = 1.02  10-7 m
 E2 = – 12.2 eV
λ3 = 0.97  10-7 m
 E3 = – 12.8 eV
λ4 = 0.94  10 m
 E4 = – 13.1 eV
λ5 = 0.93  10 m
 E5 = – 13.3 eV
λ6 = 0.93  10-7 m
 E6 = – 13.4 eV
-7
-7
y
25 m s-1
30
x
9.
The minimum energy an electron in the ground state of hydrogen can absorb is 10.2 eV. This
would allow it to jump into the n = 2 shell.
The maximum energy an electron in the ground state can absorb and stay in the atom is 13.6 eV
If it absorbs more than this it escapes the atom; any excess energy becomes kinetic energy.
10. When an electron drops from  (free) to n = 1 (ground state) it emits one 13.64 eV photon.
This would have a frequency in the U-V band of the e-m spectrum.
11. The charge of the nucleus – i.e. the number of protons it contains – determines the energy
levels. Since they all have more protons than does hydrogen, the energy levels of larger atoms
must have higher energies than the equivalent levels of hydrogen. They will therefore both
absorb and emit photons having higher energy than does hydrogen – provided the energy shells
of these atoms are compared against the equivalent energy shells of hydrogen. This implies that
photons having more energy, hence higher frequencies, would be absorbed or released by such
atoms compared to hydrogen. However their outer (valence) shell is not n = 1 as is hydrogen’s.
12. Hydrogen can neither absorb nor emit X-rays. The data in Question 4 shows that the photons of
X-rays have at least 600 eV, which would ionise the hydrogen’s solitary electron, and as has
been shown, the greatest amount of energy available from hydrogen’s electron is only13.64 eV.
13. 15 possible different photons. [n6 → n5 n6 → n4 n6 → n3 n6 → n2 n6 → n1 n5 → n4 n5 → n3
n5 → n2 n5 → n1 n4 → n3 n4 → n2 n4 → n1 n3 → n2 n3 → n1 and n2 → n1]
14. On the diagram, the Lyman series is: 121.5 nm, 102.6 nm, 97.2 nm and 95.0 nm.
The Balmer series is: 656.3 nm (red), 486.2 nm (blue-green), 434.1 nm (violet)
The Paschen series is: 1875 nm, 1282 nm.
The shortest wavelength of any series corresponds to the largest possible energy drop, from n
to n = 1, 2 or 3, depending on the series. The Lyman series converges to 91.2 nm (U-V), the
Balmer series converges to 364.6 nm (U-V) and the Paschen series to 820 nm (I-R). Only four
lines in the Balmer series are in the visible range, one red, one blue-green and one violet.
(There is one other violet line, corresponding to n6 → n2).
15. The energy levels of the element carbon in diamonds are such that no possible absorption of
photons of visible light is possible. Hence, all the visible light passes out of the diamond again,
so it is transparent. However, the elements in the impurities that make coloured diamonds must
have certain energies that correspond to specific absorptions of photons of visible light.
Absorption of both green and blue allows extra red light to pass through, so a material will
appear pink; absorption of photons corresponding to blue allows extra red and green to pass
through, resulting in what is observed as yellow, while absorption of both green and red light
allows more blue to pass through.