# 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

**(b)**

(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.

**(c)**

(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)

(ii)

Wavelength to use, Î» = 486nm

Î”E = hc / Î» = (6.63×10

^{-34}) × (3.0×10^{8}) / (4.86×10^{-9}) = 4.09×10^{-19}J (allow 2s.f)

__Question 989: [Gravitation > Weightlessness]__**(a)**The Earth may be considered to be a uniform sphere of radius 6.37 × 10

^{3}km with its mass of 5.98 × 10

^{24}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:__**(a)**

(i)

1.

Gravitational force F = Gm

_{1}m_{2}/ x^{2}
F = (6.67 × 10

^{–11}× 2.50 × 5.98 × 10^{24}) / (6.37 × 10^{6})^{2}
F = 24.6 N

2.

Centripetal force F = mxÏ‰

^{2}OR F = mv^{2}/ x**and**v = Ï‰x
Centripetal force F = 2.50 × 6.37 ×
10

^{6}× (2Ï€ / 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 F

_{G}is ‘equal’ to the centripetal force F

_{C}, the ‘weight’ / sensation of weight / contact force / reaction force which is the difference between F

_{G}and F

_{C}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 3

^{rd}law of motion. The resultant weight of the person is given by the difference in F_{G}and F_{C}. Now, when F_{G}= F_{C}, the resultant weight of the person is zero – that is the force is exerts on its surrounding is zero. Thus, from Newton’s 3^{rd}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**

*http://physics-ref.blogspot.com/2014/11/physics-9702-doubts-help-page-6.html*}

__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)

**(b)**

(i)

Î”p = 65×10

^{-3}× (-3.7 – 5.2)
Î”p = (-) 65×10

^{-3}(5.2+3.7) = (-) 0.58Ns
(ii) F = Î”p / t = 0.58 / (7.5×10

^{-3}) = 77(.3) N**(c)**

(i)

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

OR

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__would be zero. Here, the system consist of the ball of the wall.}**system**
(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.

Please consider answering ALL of the following questions:

ReplyDelete4/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)

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

Deletehttp://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-206.html

Thank you .

ReplyDeleteSolution 989:

ReplyDeletePlease explain. Doesn't weight and centripetal force act in the same direction? Towards the centre of the earth? Why aren't they added?

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

Delete