• Physics 9702 Doubts | Help Page 4

Question 22: [Radioactivity]

(a) Sample of radioactive isotope contains N nuclei at time t. At time (t + Δt), it contains (N – ΔN) nuclei of isotope. For period Δt, state, in terms of N, ΔN and Δt,

(i) Mean activity of sample

(ii) Probability of decay of nucleus

(b) Cobalt-60 source having half-life of 5.27 years is calibrated and found to have activity of 3.50 × 10⁵ Bq. Uncertainty in calibration is ±2%. Calculate length of time, in days, after calibration has been made, for stated activity of 3.50 × 10⁵ Bq to have maximum possible error of 10%.

Reference: Past Exam Paper – June 2009 Paper 4 Q9

Solution 22:

(a)

(i)

{Activity is the rate of disintegration OR the number of nuclei decaying per unit time}

Mean activity of sample = ΔN / Δt

(ii)

Probability of decay of nucleus = ΔN / N

(b)

{The calibration of the initial activity A_o contains an uncertainty of 2%. 

[Activity is the rate of disintegration OR the number of nuclei decaying per unit time] Activity decreases with time as the source decays. So, the change in activity with time tells us about the amount of nuclei decayed during that time. 

The calibration for the activity A, at time t, contains an uncertainty of 10%. So, the change in activity ΔA (= A_o – A) corresponds to the increase in error of 10 – 2 = 8%. As stated before, this percentage also tells us about the amount of nuclei that have decayed. So, the source must have decayed by 8%}

The source must decay by 8%

{Activity, A at time t = A_oexp(-λt) and the decay constant, λ = ln2 / T_½ where T_½ is the half-life.}

Activity, A = A_oexp(- (ln2) t / T_½)                  or A / A_o = 1 / (2^t/T)

{For the OR case above, - (ln2) t / T_½ = ln 2^-t/T½. So, exp(ln 2^-t/T½) = 2^-t/T½ = 1 / (2^t/T½).

The source has decayed by 8%. The initial activity (which represents the initial amount of nuclei in the source) corresponds to 100%. Therefore, the activity A (which represents the amount of nuclei at time t), at time t, corresponds to 92%. A / A_o = 92 / 100 = 0.92}

0.92 = exp(- (ln2) t / 5.27)                               or 0.92 = 1 / (2^t/5.27)

Time, t = 0.634 years = 230 days

Question 23: [Electromagnetism > Lenz’s law]

You are provided with coil of wire, bar magnet and sensitive ammeter. Outline experiment to verify Lenz’s law.

Reference: Past Exam Paper – June 2009 Paper 4 Q7

Solution 23:

A coil is connected in series with a meter [sensitive ammeter - galvanometer] (do not allow inclusion of a cell). A known pole (of a magnet) is pushed into the coil.  The current direction (not reading) is observed. The (induced) field / field from the coil repels the magnet. 

EITHER State the rule to determine the direction of the magnetic coil {Right hand grip rule – curled fingers indicate the direction of the current and the thumb, which is perpendicular to the curled finger shows the direction of the field} OR Reversing the magnet direction gives an opposite deflection on the meter.

(Lenz’s law:)The direction of the induced current is (gives rise to effects) such as to oppose the change producing it.

(The diagram was taken from the internet. Credit goes to its original owner.)

Question 24: [Radioactive decay > Momentum]

Francium-208 is radioactive and emits α-particles with kinetic energy of 1.07 × 10^–12 J to form nuclei of astatine, as illustrated.

(a) Nature of α-particle:

(b) Show that initial speed of α-particle after decay of francium nucleus is approximately 1.8 × 107ms^–1:

(c)

(i) State principle of conservation of linear momentum:

(ii) Francium-208 nucleus is stationary before decay. Estimate speed of astatine nucleus immediately after decay:

(d) Close examination of decay of francium nucleus indicates that astatine nucleus and α-particle are not ejected exactly in opposite directions. Suggest explanation for this observation:

Reference: Past Exam Paper – November 2006 Paper 2 Q3

Solution 24:

(a)

An α-particle is a helium nucleus OR contains 2 protons and 2 neutrons

(b)

{Mass of α-particle = 4u. 1u = 1.66x10^-27kg}

Kinetic energy of α-particle = ½mv²

½ [4 (1.66x10^-27)] v² = 1.07x10^-12

Initial speed of α-particle, v = 1.8x10⁷ms^-1

(c)

(i)

The principle of conservation of linear momentum states that the sum of momenta (in any direction) is constant / total momentum is constant in a closed system / no external force

(ii)

{Before decay, francium-209 nucleus is stationary – speed = 0 and hence, momentum (=mv) = 0}

Momentum of francium (= 0) = momentum of α-particle + momentum of astatine

{momentum of astatine = (-) momentum of α-particle}

204(v) = 4 (1.8x10⁷)

Speed of astatine nucleus, v = 3.5x10⁵ms^-1

(d)

{They are not ejected exactly in opposite directions – this means that the sum of momenta is not constant. So, there should be another particle / photon that would account for the missing momentum}

Another particle / photon is emitted {simultaneously} at an angle to the direction of the α-particle

Question 25: [Waves > Interference > Double Slits]

(a) State three conditions required for maxima to be formed in interference pattern produced by two sources of microwaves.

(b) Microwave source M emits microwaves of frequency 12 GHz. Show that wavelength of microwaves is 0.025 m.

(c) Two slits S₁ and S₂ are placed in front of microwave source M described in (b), as shown. Distances S₁O and S₂O are equal. Microwave detector is moved from O to P. 

Distance S₁P is 0.75 m and distance S₂P is 0.90 m. Microwave detector gives maximum reading at O. State variation in readings on microwave detector as it is moved slowly along line from O to P.

(d) Microwave source M is replaced by source of coherent light. State two changes that must be made to slits in Fig. 5.1 in order to observe interference pattern.

Reference: Past Exam Paper – June 2013 Paper 21 Q5

Solution 25:

(a)

Choose any 3:

The waves should overlap / meet / superpose

The waves must be coherent / constant phase difference (not constant λ or frequency)

The path difference = 0, λ, 2 λ OR phase difference = 0, 2π, 4π

The y must have the same direction of polarization / unpolarised

(b)

Wavelength, λ = v / f

Frequency, f = 12x10⁹Hz

Wavelength, λ = (3x10⁸) / (12x10⁹) = 0.025m

(c)

{Maxima are formed when path difference = n λ where n = 0, 1, 2, 3, … At O, the maxima corresponds to n = 0. The path difference between S₁P and S₂P = 0.90 – 0.75 = 0.15m. This corresponds to n = 0.15 / 0.025 = 6. Therefore, including the maxima at O, there is a total of 7 maxima from O to P [excluding the maxima at O and P, there is 5 maxima between O and P].

Minima are found when path difference = 0.5λ, 1.5λ, 2.5λ, … that is, midway between 2 maxima. So between 2 consecutive maxima, there is a minima. Since there is a total of 7 maxima, 6 minima are found are present between O and P.}

There is a maximum at P and several minima or maxima between O and P. There are 5 maxima / 6 minima between O and P OR 7 maxima / 6 minima including O and P.

(d)

Microwave source M is replaced by source of coherent light. State two changes that must be made to slits in Fig. 5.1 in order to observe interference pattern:

The slits must be made narrower

The slits must be put closer together

Question 26: [Electric field > Electric field strength]

Two charged points A and B are separated by distance of 6.0 cm, as shown.

Variation with distance d from A of electric field strength E along line AB is shown.

Electron is emitted with negligible speed from A and travels along AB.

(a) Relation between electric field strength E and potential V.

(b) Area below line of the graph of Fig represents potential difference between A and B. Use Fig to determine potential difference between A and B.

(c) Use answer to (b) to calculate speed of electron as it reaches point B.

(d)

(i) Use Fig to determine value of d at which electron has maximum acceleration.

(ii) Without any further calculation, describe variation with distance d of acceleration of electron.

Reference: Past Exam Paper – June 2007 Paper 4 Q3

Solution 26:

(a)

The electric field strength is equal to the negative potential gradient

{E = - V / d}

(b)

Area under the line of graph = 21.2 cm² ± 0.4 cm²

{On the x-axis (the position axis), 10 squares represent 2cm of the quantity d. But if we measure with a ruler, 1cm on the graph paper corresponds to 5 squares. So, 1cm (the length measured with ruler) on the x-axis represents [(2/10) x 5 =] 1.0cm of the quantity d. In metre this is 1.0 × 10^–2m of the quantity d.

Similarly, on the y-axis, 10 squares represent 5kVm^-1 of the quantity E. Again, if we measure with a ruler, 1cm on the graph paper corresponds to 5 squares. So, 1cm (the length measured with ruler) on the y-axis represents [(5/10) x 5 =] 2.5 kVm^-1 of the quantity E.}

 
From the graph, 1.0 cm² represents ([1.0 × 10^–2] × [2.5 × 10³] =) 25 V

Potential difference between A and B (= 21.2 x 25) = 530 V

(c)

Kinetic energy of electron, ½ mv² = qV

½ (9.1x10^-31) v² = (1.6x10^-19) (530)

Speed of electron, v = 1.37x10⁷ms^-1 

(d)

(i)

{Electric force, F = (Electric field strength, E) x (charge , q). F = ma. So, force and hence, acceleration are maximum when E is maximum}

Electron has maximum acceleration at d = 0

(ii)

The acceleration of the electron decreases then it increases.

Some quantitative analysis (e.g. it is minimum at 4.0cm)

Question 27: [Electric field > Electric potential]

(a) Define electric potential at a point.

(b) Two isolated point charges A and B are separated by distance of 30.0 cm, as shown.

Charge at A is + 3.6 × 10^–9 C. Variation with distance x from A along AB of potential V is shown.

(i) State value of x at which potential is zero.

(ii) Use answer in (i) to determine charge at B.

(c) Small test charge is now moved along line AB in (b) from x = 5.0 cm to x = 27 cm. Explain value of x at which force on test charge will be maximum.

Reference: Past Exam Paper – June 2008 Paper 4 Q4

Solution 27:

(a)

Electric potential at a point is defined as the work done moving unit positive charge from infinity to the point

(b)

(i)

Potential is zero at x = 18cm

(ii)

{At x = 18cm (this position is defined by the question), the potential is zero. So, at position x = 18cm, the sum of potential due to A and B = 0.

Note that the position is taken to be 18cm relative to A. The same position is at a distance = 30 – 18 = 12cm from B.

When calculating the potential, the distance from the point charge being considered is taken, not the values describing the positions which are only defined for convenience in the question.}

V_A + V_B = 0    {Sum of potential at x = 18cm is zero}

{Potential, V due to point where charge is q, at distance x from that point = q / 4πϵ_ox. From B, the distance is 30 – 18 = 12cm}

[(3.6x10^-9) / 4πϵ_o(18x10^-2)] + [q / 4πϵ_o(12x10^-2)] = 0

Charge, q at B = - 2.4x10^-9C

(c)

Electric field strength = (-) gradient of graph {potential gradient}

{E = - V / d and electric force, F = Eq}

Electric force = (charge, q) x gradient {potential gradient}

OR Electric force is proportional to gradient

So, electric force is largest at x = 27cm {gradient is maximum there}

Question 28: [Electric field]

(a) Two parallel metal plates P and Q are situated 8.0 cm apart in air, as shown.  Plate Q is earthed and plate P is maintained at potential of +160 V.

(i) On Fig, draw lines to represent electric field in region between plates

(ii) Show that magnitude of electric field between plates is 2.0 × 10³Vm^–1.

(b) Dust particle suspended in air between plates. Particle has charges of +1.2 × 10^–15C and –1.2 × 10^–15 C near its ends. Charges may be considered to be point charges separated by distance of 2.5 mm, as shown. Particle makes angle of 35° with direction of electric field.

(i) On Fig, draw arrows to show direction of force on each charge due to electric field.

(ii) Calculate magnitude of force on each charge due to electric field.

(iii) Calculate magnitude of couple acting on particle.

(iv) Describe the subsequent motion of particle in electric field.

Reference: Past Exam Paper – June 2005 Paper 2 Q6

Solution 28:

(a)

(i)

The lines are normal to the plates and have equal spacing (at least 4 lines). The direction is from (+) plate to earthed plate

(ii)

Electric field strength, E (= V / d) = 160 / 0.08 = 2.0x10³Vm^-1

(b)

(i)

{Electric field is from positive to negative. So, the field line shows the direction towards a negative charge. Positive charges are attracted to negatives charges and vice versa. So, the electric field line shows the direction of the force on a positive charge}

Arrows should have correct directions {force on the +ve (upper) charge is to the right [in same direction as field line] and the force on the –ve (lower) charge is to the left (since –ve charges are attracted to +ve). Note that both directions are horizontal} with the line of action of the arrows passing through the charges {each of the arrows drawn should begin at the respective charges}

(ii)

{Electric field strength, E is shown (in previous part of the question) to be 2.0x10³Vm^-1}

{Both charges have the same magnitude, so the magnitudes of the force on them are also similar}

Force, F = Eq = (2.0x10³) (1.2x10^-15) = 2.4x10^-12N

(iii)

Magnitude of couple = (one) force x perpendicular separation (of the 2 forces)

{here, the perpendicular separation is the vertical separation between the forces [which are horizontal] and equals [2.5x10^-3]sin35 }

Magnitude of couple = (2.4x10^-12) ([2.5x10^-3]sin35) = 3.4(4)x10^-15Nm

(iv)

The particle EITHER rotates {in clockwise direction} to align with the field (to become horizontal) OR oscillates (about a position) with the positive charge nearer to the earthed plate {since the other plate is +vely charged} / clockwise