IB Physics SL Theme B: The particulate nature of matter questions

Exam-style multiple choice and written questions with full markschemes. Check multiple choice answers straight away, or type, photograph or draw a written answer and get it marked.

Question 1

Multiple choiceEasy1 mark

A region of Earth is measured twice. Its albedo is unchanged between measurements; the incident and scattered powers are shown. Which row gives the albedo and the scattered power in the later measurement, respectively?

MeasurementIncident power / WScattered power / W
First600150
Later1200—
Show answer and explanation

Answer: D. Albedo is total scattered power divided by total incident power. The first measurement gives 150/600=0.25150/600=0.25 With albedo unchanged, the later scattered power is 0.25×1200 W=300 W0.25\times1200\text{ W}=300\text{ W}
A: Keeps the initial scattered power instead of scaling it for the increased incident power.
B: Adds scattered power to incident power in the denominator, giving an albedo of 150/(600+150)=0.20150/(600+150)=0.20
C: Uses the non-scattered fraction, 1−0.25=0.751-0.25=0.75 as the albedo.
Remember: multiply incident power by albedo to find scattered power.

Question 2

Paper 2Easy6 marks

Infrared radiation emitted by Earth’s surface reaches a carbon dioxide molecule in the atmosphere. Figure 1 shows two possible incoming infrared photons and two energy levels of the molecule. The molecule’s natural vibration frequency is f0f_0 In this model, a photon of frequency ff has energy E=hfE=hf

Candidate infrared photonsCO2 molecule: energy levelsA: f=f0B: f=f0/2energyE1E0ΔE=hf0
(a)

Using Figure 1, identify which photon can be absorbed by the molecule. Explain your choice using the resonance model and the energy levels.

[3]
(b)

As the molecule returns from E1E_1 to E0E_0 state the frequency of the infrared radiation emitted.

[1]
(c)

State the directions in which the radiation is emitted and explain how this can contribute to the greenhouse effect at Earth’s surface.

[2]

Question 3

Multiple choiceHard1 mark

A sample of 6.0×10236.0 \times 10^{23} particles of a monatomic gas is initially at 400 K400\ \mathrm{K} In a final state, its volume is twice its initial volume, its number density is 3/43/4 of the initial value, and its temperature is 2/32/3 of the initial value. At both states the gas is sufficiently hot and dilute for the ideal-gas approximation to apply. What is its final internal energy? Take kB=1.38×10−23 J K−1k_B=1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}}

Show answer and explanation

Answer: C. The particle-number ratio is the number-density ratio multiplied by the volume ratio: Nf/Ni=(3/4)(2)=3/2N_f/N_i=(3/4)(2)=3/2 Since U=32NkBTU=\frac{3}{2}Nk_BT the internal-energy ratio is (3/2)(2/3)=1(3/2)(2/3)=1 Thus Uf=Ui=32(6.0×1023)(1.38×10−23)(400)=4.968×103 JU_f=U_i=\frac{3}{2}(6.0\times10^{23})(1.38\times10^{-23})(400)=4.968\times10^3\ \mathrm{J} or 5.0×103 J5.0\times10^3\ \mathrm{J} to two significant figures.
A: treats the number-density ratio as the particle-number ratio, omitting the volume change.
B: accounts for the temperature change but holds the particle number fixed.
D: treats the volume ratio as the particle-number ratio, ignoring the number-density change.
Remember: for an ideal monatomic gas, internal energy depends on particle number and absolute temperature.

Question 4

Paper 2Hard10 marks

A fixed amount of gas is measured in a cylinder with a movable piston. Figure 1 shows the measured pressure–volume states of the gas at two temperatures. The temperatures are in kelvin. XX is a further state of the same gas, at an unrecorded temperature. No curves have been fitted to the measurements.

00.511.522.533.5020406080100120140160180200V / LP / kPa● 300 K■ 450 K◆ X◆
(a)

Describe the pressure–volume relationship indicated by the 300 K300\ \mathrm{K} measurements.

[1]
(b)

For the measurements at V=2.00 LV=2.00\ \mathrm{L} calculate PV/TPV/T at each temperature. Give an appropriate unit.

[2]
(c)

A fixed amount of gas changes from state 1 (P1,V1,T1)(P_1,V_1,T_1) to state 2 (P2,V2,T2)(P_2,V_2,T_2) Using the empirical pressure–temperature law at constant volume and Boyle’s law at constant temperature, derive a relation between these two states. Temperatures are in kelvin.

[2]
(d)

A controller is expected to switch when the gas reaches 750 K750\ \mathrm{K} Use state XX and both calibration results from part (b) to estimate the temperature at XX and decide whether the controller’s reported temperature is consistent with the measurements. Show your working.

[5]

Question 5

Paper 2Hard10 marks

Two samples, XX and YY have the same length and cross-sectional area. Sample XX is a metal; sample YY is an unknown material. Each sample is connected separately to a variable d.c. supply. The potential difference VV across the sample and the current II through it are measured at the same temperature.

Sample XSample YI = 20.0VI = 0.0500V0102030405060708090100110012345.00.000.050.100.150.200.250.30012345.0Current, I / mACurrent, I / µAPotential difference, V / VPotential difference, V / V
(a)

Using Figure 1, read the current in each sample when the potential difference is 3.0 V3.0\ \text{V}

[2]
(b)

Use the reading for sample XX at 3.0 V3.0\ \text{V} to determine its resistance.

[2]
(c)

A third test uses a sample SS which is either XX or YY It is connected in series with a 50 Ω50\ \Omega resistor to a 6.0 V6.0\ \text{V} supply. The current in the circuit is 60 mA60\ \text{mA} Use the circuit information and Figure 1 to determine whether SS matches XX or (Y). Show your reasoning.

[4]
(d)

A student claims that sample YY has no mobile charge carriers because its current is almost zero. Assess this claim using Figure 1 and the idea of mobility of charge carriers.

[2]

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