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

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Question 1

Multiple choiceMedium1 mark

A star and the Sun are both modelled as black bodies. The star’s surface area is four times the Sun’s, and its surface temperature is half the Sun’s; both temperatures are absolute, measured in kelvin. What is the star’s luminosity in terms of $L_{\text{sun}}$

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Answer: B. For a black body, $L=\sigma AT^4$ Comparing the star with the Sun cancels $\sigma$

\frac{L}{L_{\text{sun}}} =\frac{A}{A_{\text{sun}}}\left(\frac{T}{T_{\text{sun}}}\right)^4 =4\left(\frac12\right)^4 =\frac14.

A: includes the temperature factor but omits the area ratio.
C: uses a square rather than the fourth power of the temperature ratio.
D: includes the area ratio but ignores the temperature dependence.
Remember: for black-body radiation, luminosity depends on the fourth power of absolute temperature.

Question 2

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?

Measurement	Incident power / W	Scattered power / W
First	600	150
Later	1200	—

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Answer: D. Albedo is total scattered power divided by total incident power. The first measurement gives $150/600=0.25$ With albedo unchanged, the later scattered power is $0.25\times1200\text{ W}=300\text{ W}$
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 $150/(600+150)=0.20$
C: Uses the non-scattered fraction, $1-0.25=0.75$ as the albedo.
Remember: multiply incident power by albedo to find scattered power.

Question 3

Multiple choiceHard1 mark

A sample of $6.0 \times 10^{23}$ particles of a monatomic gas is initially at $400\ \mathrm{K}$ In a final state, its volume is twice its initial volume, its number density is $3/4$ of the initial value, and its temperature is $2/3$ 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 $k_B=1.38 \times 10^{-23}\ \mathrm{J\,K^{-1}}$

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Answer: C. The particle-number ratio is the number-density ratio multiplied by the volume ratio: $N_f/N_i=(3/4)(2)=3/2$ Since $U=\frac{3}{2}Nk_BT$ the internal-energy ratio is $(3/2)(2/3)=1$ Thus $U_f=U_i=\frac{3}{2}(6.0\times10^{23})(1.38\times10^{-23})(400)=4.968\times10^3\ \mathrm{J}$ or $5.0\times10^3\ \mathrm{J}$ 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

Multiple choiceMedium1 mark

A resistor is tested by increasing the potential difference across it and recording the current. The measured values are plotted.

Which row identifies the resistor’s behaviour and gives the ratio of power dissipated at B to that at A?

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Answer: C. At A, $V/I=2.0/0.40=5.0\ \Omega$ at B, $V/I=5.0/0.80=6.25\ \Omega$ Since the resistance is not constant, the behaviour is non-ohmic. Using $P=IV$ $P_A=0.80\ \mathrm{W}$ and $P_B=4.0\ \mathrm{W}$ so $P_B/P_A=5.0$

A: obtains the power ratio but treats the unequal $V/I$ values as equal, classifying the resistor as ohmic.
B: uses the current ratio, $0.80/0.40=2.0$ as the power ratio and omits the change in potential difference.
D: assumes constant resistance and uses $P\propto I^2$ giving a ratio of $4.0$

Remember: compare $V/I$ to identify ohmic behaviour, and use $P=IV$ for power at each operating point.

Question 5

Multiple choiceMedium1 mark

Two resistors of resistance $R$ and $2R$ are connected in series across an ideal source of potential difference $V$ They are then connected in parallel across an identical ideal source. What is the ratio of the current through the $R$ resistor in the series arrangement to the current through the $2R$ resistor in the parallel arrangement?

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Answer: B. In series, the total resistance is $3R$ so the current through $R$ is $V/(3R)$ In parallel, the $2R$ resistor has the full potential difference $V$ so its current is $V/(2R)$ The required ratio is $[V/(3R)]/[V/(2R)]=2/3$ A: uses the current through the $R$ branch in parallel, $V/R$ as the denominator. C: assumes the currents are equal because the source potential differences are equal. D: omits the $2R$ resistor when finding the series resistance, giving a series current of $V/R$ Remember: series current depends on the total resistance, while each parallel branch has the full source potential difference.

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