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, 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:
Go to 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.
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)
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
(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 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.
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.
DeleteWhy is the answer in negative in solution 990 b)i?
ReplyDeleteit depends on which direction we take as positive. It's opposite would be negative.
DeleteThe -ve sign is optional.
Ah! Thank you so much!
Deletein Q989) part 2)ii) why do we subtract centripetal force and weight?
ReplyDeleteThis comment has been removed by the author.
DeleteYour explanation here implies centripetal force acts in the opposite direction to gravitational force. Hence, both are subtracted to determine the resultant force. Can you please explain?
Deletethe explanation has been updated. see it again
Delete