Q: Applying the double-slit equation $ x = \frac{\lambda D}{a} $ to diffraction grating calculations where $ d \sin\theta = n\lambda $ must be used.

Use $ x = \frac{\lambda D}{a} $ only for double-slit interference when the angle is small. For diffraction gratings, convert lines/mm to spacing $ d $ and use $ d \sin\theta = n\lambda $.

Question 1

Using Celsius temperature instead of Kelvin in ideal gas law calculations (e.g. $ pV = NkT $ or $ pV = nRT $).

Accepted Answer

Always convert Celsius to Kelvin by adding 273.15 before performing any gas calculations (e.g. $ -10^\circ\text{C} = 263.15\text{ K} $).

Question 2

Omitting the negative sign in gravitational potential and potential energy calculations, or failing to establish the correct potential reference at infinity.

Accepted Answer

Always include the negative sign in gravitational potential formulas: $ V_g = -\frac{GM}{r} $ and $ E_p = -\frac{GMm}{r} $, acknowledging that potential is zero at infinity and negative everywhere else.

Question 3

Failing to draw a sufficiently large triangle when finding gradients from experimental data, violating practical guide tolerances.

Accepted Answer

Always draw a gradient triangle where the hypotenuse spans at least half the length of your drawn line of best fit (typically ensuring $ \Delta x \ge 17.5 $ units on the grid).

Question 4

Applying the double-slit equation $ x = \frac{\lambda D}{a} $ to diffraction grating calculations where $ d \sin\theta = n\lambda $ must be used.

Accepted Answer

Use $ x = \frac{\lambda D}{a} $ only for double-slit interference when the angle is small. For diffraction gratings, convert lines/mm to spacing $ d $ and use $ d \sin\theta = n\lambda $.

Question 5

Forgetting to convert the orbital period of geostationary satellites to seconds (24 hours = 86,400 s) when equating gravitational and centripetal forces.

Accepted Answer

Always convert time periods into SI units (seconds) before executing circular or planetary orbital equations: $ T = 24 \times 3600 = 86,400 \text{ s} $.

Question 6

Defining the work function as the energy required to release an electron, missing the word 'minimum'.

Accepted Answer

Define the work function strictly as the *minimum* energy required to release a single conduction electron from the surface of a metal.

试卷	时间	分数	题数	比重	题型
Paper 1: Modelling Physics	2小时 15分钟	100	41	37%	選擇題 (MCQ), Short Answer / Calculation, Level of Response (LoR)
Paper 2: Exploring Physics	2小时 15分钟	100	43	37%	選擇題 (MCQ), Short Answer / Calculation, Level of Response (LoR)
Paper 3: Unified Physics	1小时 30分钟	70	8	26%	Short Answer / Calculation, Level of Response (LoR)

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