Monday, January 22, 2018

The radioactive decay of a strontium (Sr) nucleus is represented in Fig. 7.1.



Question 3
The radioactive decay of a strontium (Sr) nucleus is represented in Fig. 7.1.
Fig. 7.1

(a) State whether Fig. 7.1 represents α-decay, β-decay or γ-decay. [1]

(b) One type of radioactive decay cannot be represented on Fig. 7.1.
Identify this decay and explain why it cannot be represented. [2]





Reference: Past Exam Paper – June 2007 Paper 2 Q7





Solution:
(a) β(-decay)
{Let the new element formed be X.

As the strontium (Sr) nucleus changes into X, its proton number increases to 39 while its nucleon number remains the same.

9038Sr - - > 9039X + ???

For the equation to hold, the ‘proton number’ of the radiation should be -1 such that 39 + (-1) = 38

Its nucleon number should be zero.
This corresponds to a β-particle.}


(b)
γ-decay. This is because the nucleon number and proton number of the nucleus do not change during this decay. (γ-radiation is only a loss of energy)


Thursday, January 18, 2018

A digital caliper is used to measure the 28.50 mm width of a plastic ruler. The digital caliper reads to the nearest 0.01 mm.



Question 8
A digital caliper is used to measure the 28.50 mm width of a plastic ruler. The digital caliper reads to the nearest 0.01 mm.

What is the correct way to record this reading?
A 0.02850 ± 0.01 m
B 0.0285 ± 0.001 m
C (2.850 ± 0.001) ×10-2 m
D (2.850 ± 0.001) ×10-3 m





Reference: Past Exam Paper – June 2014 Paper 13 Q6





Solution:
Answer: C.

1 mm = 10-1 cm           and      1 mm = 10-3 m

0.01 mm = 0.01×10-1 cm = 0.01×10-3 m


By writing 0.001, we need to divide 0.01 by 10. This should be removed from the 10-3 which now becomes 10-2.

So, 0.01 mm = 0.001×10-2 m


Note that when writing readings with uncertainties, the uncertainty should always be given to 1 significant figure (here it is the ‘one’ in 0.001) and the reading should be given to the same number of decimal places as the uncertainty (here, it is 3 d.p as in 2.850).

Tuesday, January 16, 2018

A particle has mass m and charge +q and is travelling with speed v through a vacuum. The initial direction of travel is parallel to the plane of two charged horizontal metal plates, as shown in Fig. 6.1.



Question 2
A particle has mass m and charge +q and is travelling with speed v through a vacuum.
The initial direction of travel is parallel to the plane of two charged horizontal metal plates, as shown in Fig. 6.1.

Fig. 6.1

The uniform electric field between the plates has magnitude 2.8 × 104 V m-1 and is zero outside the plates.

The particle passes between the plates and emerges beyond them, as illustrated in Fig. 6.1.

(a) Explain why the path of the particle in the electric field is not an arc of a circle. [1]


(b) A uniform magnetic field is now formed in the region between the metal plates. The
magnetic field strength is adjusted so that the positively charged particle passes undeviated between the plates, as shown in Fig. 6.2.
 Fig. 6.2
(i) State and explain the direction of the magnetic field. [2]

(ii) The particle has speed 4.7 × 105 m s-1.
Calculate the magnitude of the magnetic flux density.
Explain your working. [3]


(c) The particle in (b) has mass m, charge +q and speed v.
Without any further calculation, state the effect, if any, on the path of a particle that has
(i) mass m, charge –q and speed v, [1]
(ii) mass m, charge +q and speed 2v, [1]
(iii) mass 2m, charge +q and speed v. [1]





Reference: Past Exam Paper – November 2013 Paper 43 Q6





Solution:
(a) The path is not a circular arc in the field because the force on the particle acts vertically downwards.
{For circular motion, the force is towards the centre of the motion}

(b)
(i)  The direction of the magnetic field is into plane of page since the direction of the force due to the magnetic field should be opposite to that due to electric field


(ii)
{The magnitudes of the forces due to the magnetic and electric fields are equal.}
force due to magnetic field = force due to electric field
Bqv = qE
B = E / v
B = (2.8 × 104) / (4.7 × 105)
B = 6.0 × 10–2 T

(c)
{Magnetic force = Bqv                        and Electric force = Eq
For the particle to pass undeviated, Bqv = Eq
We need to understand what the changes have on both forces and which force would then be more significant.}

(i) no change / not deviated

(ii) deviated upwards

(iii) no change / not deviated
Currently Viewing: Physics Reference |