Friday, October 2, 2015

Physics 9702 Doubts | Help Page 205

  • Physics 9702 Doubts | Help Page 205

Question 988: [Quantum Physics > Energy levels in atoms]
(a) State three pieces of evidence provided by the photoelectric effect for a particulate nature of electromagnetic radiation.

(b) (i) Briefly describe the concept of a photon.
(ii) Explain how lines in the emission spectrum of gases at low pressure provide evidence for discrete electron energy levels in atoms.

(c) Three electron energy levels in atomic hydrogen are represented in Fig. 7.1.

The wavelengths of the spectral lines produced by electron transitions between these three energy levels are 486 nm, 656 nm and 1880 nm.
(i) On Fig. 7.1, draw arrows to show the electron transitions between the energy levels that would give rise to these wavelengths.
Label each arrow with the wavelength of the emitted photon.
(ii) Calculate the maximum change in energy of an electron when making transitions between these levels.

Reference: Past Exam Paper – November 2008 Paper 4 Q7

Solution 988:
(a) Choose any 3:
The ‘instantaneous’ emission (of electrons)
There is a threshold frequency below which there is no emission
The (maximum) electron energy is dependent on frequency
The (maximum) electron energy is not dependent on the intensity
The rate of emission (of electrons) depends on the intensity

(i) A photon is a ‘packet’ / quantum of energy of electromagnetic energy / radiation.

(ii) The discrete wavelengths mean that photons have particular energies. The energy of photon is determined by the energy change of an (orbital) electron. So, there are discrete energy levels.

(i) The 3 energy changes shown correctly with the arrows ‘pointing’ in the correct direction (downwards) and the wavelengths correctly identified (the smallest energy change is associated with the longest wavelength)

Wavelength to use, λ = 486nm
ΔE = hc / λ = (6.63×10-34) × (3.0×108) / (4.86×10-9) = 4.09×10-19J (allow 2s.f)

Question 989: [Gravitation > Weightlessness]
(a) The Earth may be considered to be a uniform sphere of radius 6.37 × 103 km with its mass of 5.98 × 1024 kg concentrated at its centre. The Earth spins on its axis with a period of 24.0 hours.
(i) A stone of mass 2.50 kg rests on the Earth’s surface at the Equator.
1. Calculate, using Newton’s law of gravitation, the gravitational force on the stone.
2. Determine the force required to maintain the stone in its circular path.

(ii) The stone is now hung from a newton-meter.
Use your answers in (i) to determine the reading on the meter. Give your answer to three significant figures.

(b) A satellite is orbiting the Earth. For an astronaut in the satellite, his sensation of weight is caused by the contact force from his surroundings.
The astronaut reports that he is ‘weightless’, despite being in the Earth’s gravitational field.
Suggest what is meant by the astronaut reporting that he is ‘weightless’.

Reference: Past Exam Paper – June 2015 Paper 42 Q1

Solution 989:
Gravitational force F = Gm1m2 / x2
F = (6.67 × 10–11 × 2.50 × 5.98 × 1024) / (6.37 × 106)2
F = 24.6 N

Centripetal force F = mxω2                 OR F = mv2 / x and v = ωx
Centripetal force F = 2.50 × 6.37 × 106 × (2π / 24 × 3600)2
Centripetal force F = 0.0842 N

(ii) Reading = 24.575 – 0.0842 = 24.5 N

(b) The gravitational force provides the centripetal force. When the gravitational force FG is ‘equal’ to the centripetal force FC, the ‘weight’ / sensation of weight / contact force / reaction force which is the difference between FG and FC is zero.
{As stated in the question, the sensation of weight is caused by the contact force from his surroundings. Note that this is not the usual weight W = mg that we usually use on the surface of Earth. We are talking about the ‘sensation of weight’.
For the person to feel weightless, the contact force should be zero. But this contact force is itself a consequence of Newton’s 3rd law of motion. The resultant weight of the person is given by the difference in FG and FC. Now, when FG = FC, the resultant weight of the person is zero – that is the force is exerts on its surrounding is zero. Thus, from Newton’s 3rd law, the surrounding will exert a (contact) force equal in magnitude and opposite in direction on him – and this is his ‘sensation of weight’ as stated in the question. But since the resultant weight is zero, the contact force is zero and thus, the person feels weightless.

More information on Weight – see solution 35 at}

Question 990: [Dynamics > Newton’s laws of motion]
(a) State Newton’s second law.

(b) A ball of mass 65 g hits a wall with a velocity of 5.2 m s–1 perpendicular to the wall. The ball rebounds perpendicularly from the wall with a speed of 3.7 m s–1. The contact time of the ball with the wall is 7.5 ms.
Calculate, for the ball hitting the wall,
(i) the change in momentum,
(ii) the magnitude of the average force.

(c) (i) For the collision in (b) between the ball and the wall, state how the following apply:
1. Newton’s third law,           
2. the law of conservation of momentum.

(ii) State, with a reason, whether the collision is elastic or inelastic. 

Reference: Past Exam Paper – November 2012 Paper 22 Q2

Solution 990:
(a) Newton’s second law states that the (resultant) force is equal to the rate of change of momentum                    OR change in momentum / time (taken)

Δp = 65×10-3 × (-3.7 – 5.2)
Δp = (-) 65×10-3 (5.2+3.7) = (-) 0.58Ns

(ii) F = Δp / t = 0.58 / (7.5×10-3) = 77(.3) N

1. The force on the wall from the ball is equal to the force on the ball from the wall but in the opposite direction.        

2. The momentum change of the ball is equal and opposite to the momentum change of the wall
Change of momentum of the ball and the wall is zero
{Independently, the ball and the wall has some momentum. Momentum is a vector and we need to consider its direction too. Even if the ball and the wall have some momentum, if the directions are opposite, the total momentum of the system would be zero. Here, the system consist of the ball of the wall.}

(ii) The kinetic energy (of the ball and wall) is reduced / not conserved. So, the collision is inelastic.

Question 991: [Electric field]
The diagram shows a charged particle as it approaches a pair of charged parallel plates in a vacuum.

Which row describes the horizontal and vertical components of its motion as it travels between the plates?
horizontal component             vertical component
A         constant acceleration               constant acceleration
B         constant acceleration              constant velocity
C         constant velocity                     constant acceleration
D         constant velocity                     constant velocity

Reference: Past Exam Paper – June 2011 Paper 11 Q29 & Paper 13 Q30

Solution 991:
Answer: C.
The electric field is vertical and uniform between the 2 plates. Thus, the electric force and acceleration on the charged particles would also be vertical and constant (since the field is uniform).

The experiment is done in a vacuum – air resistance would not affect the motion of the particle. Hence, the horizontal component of its velocity is constant.


  1. Please consider answering ALL of the following questions:
    4/O/N/02 Q.5(b),Q.6(c)(i)
    6/O/N/02 Q.11(a)(b)
    6/O/N/03 Q.9
    04/M/J/04 Q.8(a),(b)(i),(ii)1.
    06/M/J/04 Q.9(b)(iii),Q.11(b)
    06/O/N/04 Q.3(b)(i)1.,Q.4(a),Q.6(b),Q.8(a)(i)
    04/M/J/05 Q.7(a)
    06/O/N/05 Q.8(b),Q.10(a)
    04/M/J/06 Q.6(a),(c),Q.7(b)
    06/M/J/06 Q.14(b)
    04/O/N/06 Q.3(c)
    06/O/N/06 Q.3(b)
    05/M/J/07 Q.2(d)
    04/O/N/07 Q.7(b)(i),(c),Q.10(c)
    04/M/J/08 Q.5(b),Q.9(b),Q.11(a)(ii)
    41/O/N/09 Q.6(a),(b)(i),Q.10
    41/M/J/10 Q.6(a),Q.7(a)
    51/M/J/10 Q.2(d)

    1. For 41/M/J/10 Q.6(a), see solution 992 at

  2. Solution 989:
    Please explain. Doesn't weight and centripetal force act in the same direction? Towards the centre of the earth? Why aren't they added?

    1. Centripetal force is NOT a form of force. It is provided by other forces. Here, the weight.


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