Mixed Revision 1 19-04-17

Mixed Revision 1 – 19-04-17

Q1.The circuit diagram below shows a 6.0 V battery of negligible internal resistance connected in series to a light dependent resistor (LDR), a variable resistor and a fixed resistor, R.

(a) For a particular light intensity the resistance of the LDR is 50 kΩ. The resistance of
R is 5.0 kΩ and the variable resistor is set to a value of 35 kΩ.

(i)Calculate the current in the circuit.

current...... A

(2)

(ii)Calculate the reading on the voltmeter.

voltmeter reading ...... V

(2)

(b) State and explain what happens to the reading on the voltmeter if the intensity of the light incident on the LDR increases.

......

(2)

(c) For a certain application at a particular light intensity the pd across R needs to be 0.75 V. The resistance of the LDR at this intensity is 5.0 kΩ.

Calculate the required resistance of the variable resistor in this situation.

resistance ...... Ω

(3)

(Total 9 marks)

Q2.(a) (i)Draw and label suitable apparatus required for measuring the Young modulus of a material in the form of a long wire.

(ii)List the measurements you would make when using the apparatus described in part (i).

......

(iii)Describe briefly how the measurements listed in part (ii) would be carried out.

......

(iv)Explain how you would calculate the Young modulus from your measurements.

......

(13)

(b) A uniform heavy metal bar of weight 250 N is suspended by two vertical wires, supported at their upper ends from a horizontal surface, as shown.

One wire is made of brass and the other of steel. The cross-sectional area of each wire is 2.5 ×10–7 m2 and the unstretched length of each wire is 2.0 m.

the Young modulus for brass = 1.0 × 1011 Pa
the Young modulus for steel = 2.0 × 1011 Pa

(i)If the tension, T, in each wire is 125 N, calculate the extension of the steel wire.

......

(ii)Estimate how much lower the end A will be than the end B.

......

(3)

(Total 16 marks)

Q3. (a) When free electrons collide with atoms in their ground state, the atoms can be excited or ionised.

(i) State what is meant by ground state.

......

(1)

(ii)Explain the difference between excitation and ionisation.

......

(3)

(b) An atom can also become excited by the absorption of photons. Explain why only photons of certain frequencies cause excitation in a particular atom.

......

(4)

(c) The ionisation energy of hydrogen is 13.6 eV. Calculate the minimum frequency necessary for a photon to cause the ionisation of a hydrogen atom. Give your answer to an appropriate number of significant figures.

answer ...... Hz

(4)

(Total 12 marks)

Q4.The figure below shows a spectrometer that uses a diffraction grating to split a beam of light into its constituent wavelengths and enables the angles of the diffracted beams to be measured.

(a) Give one possible application of the spectrometer and diffraction grating used in this way.

......

(1)

(b) (i)When the spectrometer telescope is rotated from an initial angle of zero degrees, a spectrum is not observed until the angle of diffraction θ is about 50°. State the order of this spectrum.

......

(1)

(ii)White light is directed into the spectrometer. Light emerges at A and B. State one difference between the light emerging at B compared to that emerging at A.

......

(1)

(c) The angle of diffraction θ at the centre of the observed beam B in the image above is 51.0° and the grating has 1480 lines per mm.

Calculate the wavelength of the light observed at the centre of beam B.

wavelength ...... m

(3)

(d) Determine by calculation whether any more orders could be observed at the wavelength calculated in part (c).

(2)

(Total 8 marks)

M1.(a) (i)(use of I = V / R)

first mark for adding resistance values 90 k Ω

I = 6.0 / (50 000 + 35 000 + 5000) = 6.7 × 10−5A

accept 7 × 10−5 or dotted 6 × 10−5
but not 7.0 × 10−5 and not 6.6 × 10−5

(ii)V = 6.7 × 10−5 × 5000= 0.33 (0.33 − 0.35) V
OR
V = 5 / 90 × 6 = 0.33( V)

CE from (i)
BALD answer full credit
0.3 OK and dotted 0.3

(b) resistance of LDR decreases

need first mark before can qualify for second

reading increase because greater proportion / share of the voltage across R OR higher current

(c) I = 0.75 / 5000 = 1.5 × 10−4(A)
(pd across LDR = 0.75 (V))
pd across variable resistor = 6.0 − 0.75 − 0.75 = 4.5 (V)
R = 4.5 / 1.5 × 10−4 = 30 000 Ω
or
I = 0.75 / 5000 = 1.5 × 10−4 (A)
RtotalI = 6.0 / 1.5 × 10−4 = 40 000 Ω
R = 40 000 − 5000 − 5000 = 30 000 Ω

[9]

M2.(a) (i)diagram to show:
(long) wire fixed at one end (1)
mass / weight at other end (1)
measuring scale (1)
mark on wire, or means to measure extension (1)

max 3

[alternative for two vertical wires:
two wires fixed to rigid support (1)
mass / weight at end of one wire (1)
other wire kept taut (1)
spirit level and micrometer or sliding vernier scale (1)]

(ii)measurements:
length of the wire between clamp and mark (1)
diameter of the wire (1)
extension of the wire (1)
for a known mass (1)

max 3

(iii)length measured by metre rule (1)
diameter measured by micrometer (1)
at several positions and mean taken (1)
(known) mass added and extension measured
by noting movement of fixed mark against vernier scale
(or any suitable alternative) (1)
repeat readings for increasing (or decreasing) load (1)

max 5

(iv)graph of mass added / force against extension (1)

gradient gives (1)

correct use of data in where A is cross-sectional area (1)

[if no graph drawn, then mean of readings
and correct use of data to give 2max) (1)

max 2

(13)

The Quality of Written Communication marks are awarded for the quality of answers to this question.

(b) (i)for steel (use of gives) e = (1)

e = (1)

= 5.0 × 10–3 m (1)

(ii)extension for brass would be 10 × 10-3(m) (or twice that of steel) (1)
end A is lower by 5 mm √ (allow C.E. from (i))

max 3

[16]

M3.(a) (i)when electrons/atoms are in their lowest/minimum energy (state) or
most stable (state) they (are in their ground state)

(ii)in either case an electron receives (exactly the right amount of) energy

excitation promotes an (orbital) electron to a higher energy/up a level

ionisation occurs (when an electron receives enough energy) to leave
the atom

(b) electrons occupy discrete energy levels

and need to absorb an exact amount of/enough energy to move to a higher level

photons need to have certain frequency to provide this energy or e = hf

energy required is the same for a particular atom or have different energy levels

all energy of photon absorbed

in 1 to 1 interaction or clear a/the photon and an/the electrons

hf = 2.176 × 10−18

f = 2.176 × 10−18 ÷ 6.63 × 10−34 = 3.28 × 1015 Hz 3 sfs

[12]

M4.(a) one of:
(spectral) analysis of light from stars
(analyse) composition of stars
chemical analysis
measuring red shift \ rotation of stars ✓

insufficient answers:
‘observe spectra’, ‘spectroscopy’, ‘view absorption \ emission spectrum’, ‘compare spectra’, ‘look at light from stars’.

Allow : measuring wavelength or frequency from a named source of light

Allow any other legitimate application that specifies the source of light. E.g.
absorbtion \ emission spectra in stars,
‘observe spectra of materials’

(b) (i)first order beam
first order spectrum
first order image
✓

Allow ‘n = 1’ ,‘1’ , ‘one’, 1st

(ii)the light at A will appear white (and at B there will be a spectrum)
OR greater intensity at A ✓

( nλ = d sin θ )
= 6.757 × 10−7 × sin 51.0 ✓ecfonlyfor :
•incorrect power of ten in otherwise correct calculation of d
•use of d = 1480, 1.48, 14.8 (etc)
•from incorrect order in bii

= 5.25 × 10−7 (m) ✓ecfonlyfor :
•incorrect power of ten in otherwise correct d
•from incorrect order in bii

Some working required for full marks. Correct answer only gets 2

Power of 10 error in d gets max 2

For use of d in mm, answer =
5.25 × 10−4 gets max 2

n = 2 gets max 2 unless ecf from bii

use of d = 1480 yields wavelength of 1150m

(d) n = d (sin90) / λOR n = 6.757 × 10−7 / 5.25 × 10−7✓ecf both numbers from c

= 1.29 so no more beams observed ✓or answer consistent with their working

2 = d (sinθ) / λ OR sinθ = 2 × 5.25 × 10−7 / 6.757 × 10−7✓ecf both numbers from c

sinθ = 1.55 (so not possible to calculate angle) so no more beams ✓

OR sin−1(2 × (their λ / their d) ) ✓
(not possible to calculate) so no more beams ✓ecf

Accept 1.28, 1.3

Second line gets both marks

Conclusion consistent with working

[8]

E1.This question on a potential divider circuit was a mixture of qualitative and quantitative. As is often the case with questions involving electric circuits, candidates coped better with quantitative parts. This was particularly true in part (a) where the calculation involved more than one stage.

Part (b) was not well done and only the strongest candidates manage to relate the changing light intensity to the voltmeter reading. A significant proportion of candidates were under the impression that increasing the light intensity increases the ldr resistance.

Part (c) did involve a calculation but this was much more challenging than part (a) because there were no intermediate stages. Only a third of candidates were able to calculate a correct value for the resistance of the variable resistor. The majority of those who were successful calculated the value using a ratio method rather than calculating the current and then using this value with the correct pd to find the resistance.

E2.Almost all candidates gained reasonable marks on part (a)(i) even though some of the descriptions were lacking in detail. Most of the diagrams were reasonably drawn with the aid of a ruler. Candidates who drew freehand usually produced inferior diagrams which failed to gain all the available marks. A variety of methods were shown, usually two wires hanging vertically, linked together by means of some vernier arrangement and a spirit level. Also shown was a horizontal wire on a bench. Although this is not such an accurate method, it was accepted but many candidates showed the mark on the wire as being about half way along. This of course is only acceptable if the length of the wire is measured to that point, but this was usually overlooked in the description. The least satisfactory method was suspending a single wire with a ruler alongside although this did gain some marks. There were an alarming number of diagrams which showed a completely unrelated length of wire, a ruler, an isolated hook with a mass attached and a micrometer. Needless to say, such efforts gained no marks.

In part (ii) candidates could have saved themselves considerable time and effort by reading the question carefully and just listing the measurement they would make. Many candidates listed the area of cross section as a measurement. This was not acceptable since area is a derived quantity and it is the diameter which is measured. Many candidates also listed the ‘width’ of the wire, which again was not accepted.

The descriptions in part (iii) were, on the whole, quite reasonable, although most effort seemed to go into describing how the length of the wire and its diameter were measured and not giving sufficient attention to the experiment, i.e. measuring the extension for each mass added and increasing the total mass to a certain value. There were very few references to repeating the readings while unloading. This particular section of the question was also used to award the quality of written communication marks and most candidates scored well on this.

The descriptions in part (iv) of how to use the measurements to give the Young modulus was reasonably done with about 50% of the candidates drawing a graph of force vs extension or stress vs strain and using the gradient accordingly. Candidates who only used one set of values to give one value of the Young modulus were not awarded all the available marks.

The calculation in part (b)(i) was performed satisfactorily, with the majority of candidates calculating the correct extension for the steel wire. Marks were lost in part (ii) when the answer was given without any reasoning.

E3. Many students were able to distinguish between excitation and ionisation successfully and also to define the ground state. They clearly found the structured format of this question helpful. However, students were not so good at explaining the process of excitation of atoms by the absorption of photons. It was common to see muddled answers that confused the photoelectric effect with excitation. The term work function was often used incorrectly in candidate responses as was threshold frequency. A significant minority focused on the photon released after excitation rather than the incident photon.

The calculation in part (c) was generally done well and most students gave answers to the correct number of significant figures. A common error by some students was to fail to convert electron volts to joules, this mistake limited them to a maximum of two marks.

E4.(a)There were some rather vague answers here such as ‘To calculate the wavelength of a light’ or ‘to look at the light from stars’. There needed to be a little more than this to get the mark, i.e. a specific example such as ‘analyse the elements present in the atmosphere of a star’ or explain that the composition of a material or gas can be determined.

(b) The candidates who knew this often lacked detail in their answer, e.g. ‘it would be dimmer’. Some thought there would only be one colour at B rather than a spectrum.

Quite a few thought that the wavelength at B would be different from A due to the increased angle.

Some candidates thought that the light at B would be composed of different wavelengths and the white light at A would be a single wavelength.

(c) This was a fairly standard exam question but surprisingly there were few correct answers. Students seemed to be poorly prepared for this question and confusion reigned regarding the meaning of the terms in the grating equation. Use of the lines per mm as the line spacing (d = 1480) was very common.

There was also confusion between line spacing, d, and order, n. Some used 1480 for d and for n.

Candidates often used 1 / 1480 and then failed to convert this into metres.

(d) There were a surprising number of candidates who did not attempt this question.
Even if they felt they had the wrong numbers for wavelength and line spacing in part (c), candidates simply needed to divide their d by their λ, and if greater than 1, conclude that no further orders are possible.

There was also some confusion over the method required, e.g. some used the angle given in part (c), (51°), and calculated a new wavelength that would give a second order at that angle.

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