Particles And Waves Flashcards
9.) D
In beta minus decay, a neutron transforms into a proton, an electron and an antineutrino.
In beta plus decay, a proton transforms into a neutron, a positron, and a neutrino.
10.) B
Ek = eV
Ek = 1/2 mv^2
eV = 1/2 mv^2
Now rearrange to find v
1.) multiply both sides to get rid of the fraction.
2eV = mv^2
2.) divide by m to isolate v^2
2eV / m = v^2
3.) square root it
√ 2eV / m = v
v = √ 2eV / m
11.) A
If the electron starts with a higher speed it will spend less time between the plates and experience less deflection. This will allow it to reach the screen.
Increasing the potential difference means a stronger electrical field which increases the force on the electron which would cause it to miss the screen by an even larger distance.
Reversing the polarity of the plates would cause the electron to move downwards instead of upwards. Since the electron doesn’t reach the plate originally we know that if the polarity is reversed it will still travel the same distance just in a downwards trajectory instead and still not reach the plate.
12.) D
E = Δmc^2
E = (6.684x10^-27 - 6.68x10^-27) x (3x10^8)^2
E = 3.6x10^-13 J
13.) D
v = f λ
f = v / λ
f = 340 / 0.3
f = 1133 Hz
14.) D
15.) A
A smaller critical angle means a higher refractive index.
Since material P has a higher refractive index the wavelength of the red light will be shorter.
16.) E
Higher energy transitions correspond to shorter wavelengths.
The largest energy difference corresponds to the transition with the greatest drop in energy levels.
6a.) photon
B.)
E = 1.6 x 10 ^-19 x 126
= 2.016 x 10 ^-17
E = 2.016 x 10 ^-17 x 10^9
E = 2 x 10 ^-8
E = mc^2
m = E / c^2
m = (2 x 10^-8) / (3 x 10^8)^2
m = 2.2 x10^-25
iii.) 10^-25 / 10^-27 = 100. Therefore a Higgs boson is 2 orders of magnitude bigger
8a.) the power per unit area
b.) 134 × 0·22^2 = 5·4
60·5 × 0·32^2 = 5·4
33·6 × 0·42^2 = 5·4
21·8 × 0·52^2 = 5·5
Statement of I × d^2= constant
C.)
I x d^2 = 5.4
I x 0.60^2 = 5.4
I = 5.4 / 0.60^2
I = 15 Wm ^-2
D.)
smaller lamp will be more like a point source.
Black cloth on bench to reduce reflections.
Use a more precise instrument to reduce absolute uncertainty.
E.)
A=4πr^2
=4πx2^2 =50·265
I = P / A
I = 24 / 50·265
I = 0·48 W m^-2
9ai.)
Different frequencies/ colours have different refractive indices
or
Different frequencies/ colours are refracted through different angles
ii.)
n = v1 / v2
v2 = v1 / n
v2 = 3.0 x 10^8 / 1.54
v2 = 1.95 x 10^8 ms^-1
9bi.)
v = f λ
λ = v / f
λ = 3.0 x 10^8 / 4.57 x 10 ^14
λ = 6.57 x10^-7
d sin θ = m λ
d sin 19 = 2 x 6.57 x 10^-7
d sin 19 = 1.314 x10^-6
d = 1.314 x 10^-6 / sin 19
d = 1.314 x 10^-6 / 0.325568
d = 4 x 10^-6 m
ii.)
different colours have different λ
mλ = d sinθ
(m and d are the same)
θ is different for different λ
8.) A
Given:
1 up quark (charge = +⅔)
2 down quarks (each charge = −⅓)
Total charge =
= (+⅔) + (−⅓) + (−⅓)
= 0
Type of hadron:
Hadrons are either baryons (3 quarks) or mesons (1 quark + 1 antiquark).
Since we have three quarks, it is a baryon.
Correct row: A
• Charge: 0
• Type: baryon
9.) E
I. The force-mediating particles are bosons → True
Bosons include photons, gluons, W/Z bosons, etc.
II. Gluons are the mediating particles of the strong force → True
III. Photons are the mediating particles of the electromagnetic force → True
10.) D
We’re told:
• A Bismuth (Bi) nucleus undergoes beta decay, forming Polonium (Po).
• That Polonium nucleus then undergoes alpha decay, forming Lead (Pb) with:
• Mass number = 208
• Atomic number = 82
We are to find the missing mass numbers and atomic numbers (P, Q, R, S) for each step in the decay series:
Step 1: Alpha Decay (Po → Pb)
Alpha decay emits a helium nucleus:
• Mass number decreases by 4
• Atomic number decreases by 2
So:
• Mass number of Po = 208 + 4 = 212 → So R = 212
• Atomic number of Po = 82 + 2 = 84 → So S = 84
Step 2: Beta Decay (Bi → Po)
Beta decay converts a neutron to a proton:
• Mass number stays the same
• Atomic number increases by 1
So:
• Mass number of Bi = 212 → So P = 212
• Atomic number of Bi = 84 − 1 = 83 → So Q = 83
Final Values:
• P = 212
• Q = 83
• R = 212
• S = 84
Correct Answer: D
In alpha decay mass number decreases by 4 ( 2 protons & 2 neutrons).
The atomic number decreases by 2 (2 protons)
In beta decay mass number stays the same.
Atomic number increases by 1.
11.) D
12.) A
Given:
• Frequency, f = 9.00 × 10¹⁵ Hz
• Maximum kinetic energy, Ek = 5.70 × 10⁻¹⁸ J
• Find: Work function, φ
Use the photoelectric equation:
E = φ + Ek
Where:
• E = hf, and h = Planck’s constant = 6.63 × 10⁻³⁴ J·s
Calculate E = hf:
E = 6.63 × 10⁻³⁴ x 9.00 × 10¹⁵
E = 5.967 x 10⁻¹⁸
Now solve for φ:
E = φ + Ek
φ = E - Ek
φ = 5.967 x 10⁻¹⁸ - 5.70 × 10⁻¹⁸
φ = 2.67 × 10⁻¹⁹ J
13.) B
d sin θ = m λ
d = m λ / sin θ
d = 1 x 633 × 10⁻⁹ / 0.3090169944
d = 2.05 × 10⁻⁶ m
14.) E
When light moves from glass (denser) into air (less dense):
• Speed increases
• Frequency stays the same (frequency is determined by the source)
• Wavelength increases (since speed increases and f is constant, λ must increase)
15.)
I = k / d²
k = I x d²
k = 32 x 4²
k = 512
I = k / d²
I = 512 / 16²
I = 2.0 W/m²
16.) B
What’s in the diagram:
• X: Transition from E₂ to E₁ (smaller energy change)
• Y: Transition from E₂ to E₀ (larger energy change)
Now look at the statements:
Statement I:
Transition Y produces photons of higher frequency than transition X. • Since Y has a greater energy difference than X, it produces photons with more energy.
✅ This statement is correct.
Statement II:
Transition X produces photons of longer wavelength than transition Y. • Again, since X has less energy, it has lower frequency.
→ Lower frequency = longer wavelength
✅ This statement is also correct.
Statement III:
When an electron is in the energy level E₀, the atom is ionised. • E₀ is the lowest energy level — it’s the ground state, not ionisation. • An atom is ionised when the electron is removed completely, i.e. energy = 0 or higher (above the energy levels shown).
❌ This statement is incorrect.
7a.)
Ek = QV
Ek = (1.60 x 10^-19) x ( 2 x 10^3)
Ek = 3.2x10^-16 J
8.a) mass is converted into energy.
B.)
M reactants = 3.3436 x 10^-27 +
5.0083 x 10 ^-27
M reactants = 8.3519 x 10^-27kg
M products = 6.6465 x 10^-27 +
1.6749 x 10^-27
M products = 8.3214 x 10^ -27kg
Δm = (8.3519 x 10^-27) - 8.3214 x 10^-27
Δm = 3.05 x 10^-29kg
E = mc^2
E = 3.05 x 10^ -29 x (3x10^8)^2
E = 2.745 x 10^ -12
8c.)
Plasma would cool down if it came too close to the sides and the reaction would stop.
OR
Reaction requires very high temperature so plasma would melt the sides of the reactor.
OR
High Temperature plasma could destroy the container
D.) upwards
9.a) the waves from the two sources have a constant phase relationship and have the same frequency, wavelength and velocity.
B.)
Waves meet in phase OR Crest meets crest
OR Trough meets trough OR Path difference = m λ
C.)
Path difference = m λ
λ = path difference / m
λ = (282x10^-3 - 204x10^-3 ) / 2
λ = 0.039m
9d.) the path difference stays the same because the wavelength has not changed.
10a.)
n = sin θ1 / sin θ2
n = sin 36 / sin 18
n = 0.5877852523 / 0.3090169944
n = 1.9
10b.)
Sin θc = 1 / n
Sin θc = 1 / 1.9
Sin θc = 0.5
θc = 32 degrees
C.) Completed diagram, showing light emerging (approximately) parallel to the incident ray
8.) A
9.) D
10.) B
E = hf
Work function = E - Ek
Work function = hf - Ek
Work function = 6.63x10^-34 x 7.70x10^14 - 2.67x10^-19
Work function = 2.4351 x 10^-19 J
11.) E
