1.2 - Basic Ideas About Atoms

Question 1

(a)

Atomic structure

Answer 1

Proton:
relative mass 1, relative charge +1

Neutron: relative mass 1, relative charge 0

Electron:
relative mass negligible, relative charge -1

Question 2

(a)

Relative atomic mass

Answer 2

number of protons

Question 3

(a)

Mass number

Answer 3

sum of protons and neutrons

Question 4

(a)

Isotopes

Answer 4

Atoms of the same element with different numbers of neutrons.

Question 5

(a)

Alpha (α) particles

Answer 5

positively charged helium nuclei, mass of four units, stopped by a piece of paper, strongly ionising

Question 6

(a)

Beta (β) particles

Answer 6

Negatively charged electrons, negligible mass, stopped by 0.5 cm of aluminium

Question 7

(a)

Gamma (γ) radiation

Answer 7

very high energy electromagnetic
radiation, > 2cm of lead required to stop it, weakly ionising.

Question 8

(a) Equations

Alpha decay (α)

Answer 8

A helium nucleus is produced during this decay, so the atomic number of the original element decreases by two, and the mass number decreases by four.

Question 9

(a)

Alpha decay example
(238,92) U → (,) Th + (,) He

Answer 9

(238,92) U → (234,90) Th + (4,2) He

Question 10

(a) Equations

Beta decay (β)

Answer 10

The atomic number of the original
element increases by one, and has the same mass number

Question 11

(a)

Beta Decay Example
(14,6) C → (,) N + (,) β

Answer 11

(14,6) C → (14,7) N + (0,-1) β

Question 12

(a) Equations

Positron emission (β+
decay)

Answer 12

The atomic number of the original
element decreases by one, and
has the same mass number.

Question 13

(a)

Positron emission example
(11,6) C → (,) B + (,) β

Answer 13

(11,6) C → (11,5) B + (0,1) β

Question 14

(a) Equations

Electron capture (inverse β decay)

Answer 14

The atomic number of the original element decreases by one, forming a different element and has the
same mass number.

Question 15

(a)

Electron capture example (41,20) Ca → (,) e + (,) K

Answer 15

(41,20) Ca → (0,-1) e + (41,19) K

Question 16

(b)

behaviour of α-radiation in electric and magnetic field and relative penetrating power

Answer 16

α-particles are positive, heavy and slow moving and are attracted slightly to the negative plate
of an electric field.

Question 17

(b)

behaviour of β-radiation in electric and magnetic field and relative penetrating power

Answer 17

β-particles are light and fast moving and show considerable deviation towards the positive plate of an electric field.

Question 18

(b)

behaviour of γ-radiation in electric and magnetic field and relative penetrating power

Answer 18

γ-radiation is electromagnetic radiation of short wavelength and is unaffected by an electric
field.

Question 19

(c)

Half-life of radioactive decay

Answer 19

The half-life of a radioisotope is the time taken for its radioactivity to fall to half of its initial value

Question 20

(c) Half-life calculations

Ra-226 has a half-life of 1600 years.
How many years will it take to decay to 1/16 of its original value?

Answer 20

1 → 1/2 → 1/4 → 1/8 → 1/16
Four half-lives so 4 × 1600 6400 years

Question 21

(c) Half-life calculations

Ra-226 has a half-life of 1600 years.
If the initial mass of Ra-226 was 20.0g, what mass would be
remaining after 4800 years?

Answer 21

4800/1600 = 3 half-lives
20.0 →10.0 →5.0 →2.5
The mass halves three times 2.5 g remains

Question 22

(d)

Adverse consequences for living cells of exposure to radiation

Answer 22

Ionising radiation can damage the DNA of a cell. Damage to the
DNA may lead to changes in the way the cell functions, which can
cause mutations and the formation of cancerous cells at lower
doses or cell death at higher doses. Exposure to high levels of
radiation can cause radiation burns and death.

Question 23

(d)

Use of
radioisotopes

Answer 23

Treatment of cancer
Tracer
Calculating age of plant and animal remains
Production of electricity.
Measuring the thickness of metal foil.

Question 24

(e)

Ionisation energy

Answer 24

The energy required to remove one mole of electrons from one mole of atoms in the gaseous state.

Question 25

(e)

Factors that influence the attraction of the ionisation energy

Answer 25

• the size of the positive nuclear charge
• the distance of the outer electron from the nucleus
• the shielding effect of electrons in fully occupied inner shells.

Question 26

(e)

Trend in ionisation energy across a period

Answer 26

Ionisation energy increases across a period because the number of protons increases and electron shielding remains very similar. This means across the period the electrons require more energy to overcome the strengthening nuclear attraction.

Question 27

(e)

Trend in ionisation energy down a group

Answer 27

Ionisation energy decreases down a group. This is because the atomic radius and electron shielding increases so the nuclear attraction with the electron gets increasingly weaker making the electron easier to remove.

Question 28

(f)

Link between successive ionisation energy values and electronic structure

Answer 28

Successive ionisation energies always increase because:
* there is a greater ‘effective’ nuclear charge as the same number
of protons are holding fewer and fewer electrons
* as each electron is removed, each shell will be drawn slightly
closer to the nucleus
* as the distance of each electron from the nucleus decreases,
the nuclear attraction increases.

Question 29

(g)

s orbital

Answer 29

• spherical
• can hold up to two electrons

Question 30

(g)

d orbital

Answer 30

• five different orbitals
• can hold up to 10 electrons in total (two in each orbital)

Question 31

(g)

p orbital

Answer 31

• dumb-bell shaped lobes at right angles
• can hold up to six electrons in total (two in each orbital)

Question 32

(g)

order of blocks

Answer 32

1s,2s,2p,3s,3p,4s,3d,4p,5s,4d,5p,6s,5d,6p,7s,6d

Question 33

(h)

Origin of emission and absorption spectra in terms of electron transitions between atomic energy levels

Answer 33

When an electron moves down an energy level, it releases a photon to conserve energy. Conversely, when an electron gains energy to move up an energy level, it absorbs a photon. The specific wavelengths and energies of these photons are observed in emission and absorption spectra.

Question 34

(i)

Atomic emission spectrum of the hydrogen atom

Answer 34

Emission spectra - displays lines at the specific frequencies of emitted photons.

Absorption spectra - displays an entire spectrum with black lines for the ‘missing’ frequencies of the absorbed photons.

Question 35

(j)

Relationship between energy and frequency

Answer 35

(E = hf)
Energy (J) = Planck’s constant (m2 kgs-1) x Frequency of photon (Hz)

Question 36

(j)

Relationship between frequency
and wavelength

Answer 36

(f = c/λ)
Frequency (Hz) = Speed of light (m/s)/ Wavelength (m)

Question 37

(j)

Equation between energy, frequency and wavelength

Answer 37

E= h (c/λ)

Question 38

(h)

Order of increasing energy

Answer 38

Infrared < visible < ultraviolet light

Question 39

(l)

Significance of the frequency of the convergence limit of the Lyman series and its relationship with the ionisation energy of the hydrogen atom

Answer 39

The frequency of the convergence limit of the Lyman series can be used in the equation E=hv to calculate the first ionisation energy for hydrogen.