Physics 9702 Doubts | Help Page 220
Question 1048: [Applications
> Communicating information]
(a)
State two reasons why frequencies in the gigahertz (GHz) range are used in
satellite communication.
(b)
In one particular satellite communication system, the frequency of the signal
transmitted from Earth to the satellite (the up-link) is 6 GHz. The frequency
of the signal transmitted back to Earth from the satellite (the down-link) is 4
GHz.
Explain why the
two signals are transmitted at different frequencies.
(c)
A signal transmitted from Earth has a power of 3.1 kW.
This signal,
received by a satellite, has been attenuated by 185 dB.
Calculate the
power of the signal received by the satellite.
Reference: Past Exam Paper – November 2013 Paper 43 Q12
Solution 1048:
(a) e.g.
no / little
ionospheric reflection
large
information carrying capacity
(b)
This prevents the
(very) low power signal received at satellite from being swamped by the high-power
transmitted signal
(c)
Attenuation /
dB = 10 lg(P2/P1)
{P2
= power of signal transmitted from Earth and P1 = power of (attenuated)
signal received by satellite}
185 = 10
lg({3.1×103} / P)
P = 9.8 × 10–16
W
Question 1049: [Matter > Thermal properties of materials >
Specific latent heat]
(a) Define specific latent heat of fusion.
(b) Some crushed ice at 0 °C is placed in a funnel together with an
electric heater, as shown in Fig. 2.1.
The mass of water collected in the
beaker in a measured interval of time is determined with the heater switched
off. The mass is then found with the heater switched on. The energy supplied to
the heater is also measured.
For both measurements of the mass,
water is not collected until melting occurs at a constant rate.
The data shown in Fig. 2.2 are
obtained.
(i) State why the mass of water is
determined with the heater switched off.
(ii) Suggest how it can be
determined that the ice is melting at a constant rate.
(iii) Calculate a value for the
specific latent heat of fusion of ice.
Reference: Past Exam Paper – November 2008 Paper 4 Q2
Solution 1049:
(a) Specific latent heat of fusion is defined as (thermal) energy /
heat required to convert unit mass of solid to liquid at its normal melting
point / without any change in temperature.
(b)
(i) To make allowance for
heat gains from the atmosphere
(ii) Choose any 1:
By the constant rate of production
of droplets from the funnel
When a constant mass of water is
collected per minute in the beaker
(iii)
{Even with the heater off,
some of the ice would still melts, forming water. In 10.0min 16.6g of water has
been formed. So, in 5.0min half that amount of water is formed.
To calculate the specific
latent heat of ice, we need to consider the amount of ice melted only due to
the heater being switched on. So, the amount of water formed without the heater
being on should be subtracted from the total mass of water formed in 5.0min
with the heater being on.}
Mass melted by heater in 5 minutes =
64.7 – (½ × 16.6)
= 56.4g
{Mass × Specific latent heat of fusion = Amount of heat energy supplied to
the heater}
56.4×10-3 × L = 18 kJ
Specific latent heat of fusion, L =
320 kJ kg-1
Question 1050 [Electric field > Electric potential]
Two point charges A and B each have
a charge of + 6.4 × 10–19 C. They are separated in a vacuum by a
distance of 12.0 μm, as shown in Fig. 4.1.
Points P and Q are situated on the
line AB. Point P is 3.0 μm from charge A and point Q is 3.0 μm from charge B.
(a) Calculate the force of repulsion between the
charges A and B.
(b) Explain why, without any calculation, when a
small test charge is moved from point P to point Q, the net work done is zero.
(c) Calculate the work done by an electron in
moving from the midpoint of line AB to point P.
Reference: Past Exam Paper – June 2010 Paper 42 & 43 Q4
Solution 1050:
(a)
Force = q1q2 /
4πϵox2
Force = (6.4×10-19)2
/ [4π × (8.85×10-12) × (12×10-6)2]
Force = 2.56×10-17N
(b)
{Electric potential, V = q
/ 4πϵor
The electric potential V
at a point is the work done in bringing unit positive charge from infinity to
that point.
Consider a charge at point
P. It is affected by the field of 2 charges: a charge of + 6.4 × 10–19
C at a distance of 3.0μm (located at A) and a charge of + 6.4 × 10–19
C at a distance of 9.0μm (located at B).
Now consider a charge at
point Q. It will be affected in the field in a similar way, except that the
charge at A is at a distance of 9.0μm and the charge at B is at a distance of 3.0μm.
Thus, P and Q are at the
same electric potential.}
The potential at P is the same as
the potential at Q. The work done = qΔV. Since ΔV = 0, the net work done is
zero.
(c)
{V = q / 4πϵor.
At midpoint, potential = VA + VB. The distance r is 6.0μm.
Since both potentials are the same, we can take twice the value of one of the
potentials.}
At the midpoint, potential = 2 ×
[(6.4×10-19) / 4πϵo (6×10-6)]
{At point P, the value of
r in VA = 3.0μm and the value of r for VB is 9.0μm.}
At point P, potential = [(6.4×10-19)
/ 4πϵo (3×10-6)] + [(6.4×10-19) / 4πϵo×(9×10-6)]
{At the midpoint,
potential = 2×[(6.4×10-19) / 4πϵo (6×10-6)] = (6.4×10-19)
/ 4πϵo (3×10-6)
When taking the difference
in potentials at P and at the midpoint, this terms ‘(6.4×10-19) / 4πϵo
(3×10-6)’ would get eliminated.}
Change in potential (= potential at P – potential at midpoint) = (6.4×10-19)
/ 4πϵo×(9×10-6)
Energy (= qV)
= (1.6×10-19) × [(6.4×10-19) / 4πϵo×(9×10-6)]
{q is the charge of the
electron.}
Energy = 1.0×10-22J
Question 1051: [Quantum
Physics > Line Spectra]
White light is incident on a cloud
of cool hydrogen gas, as illustrated in Fig. 8.1.
The spectrum of the light emerging
from the gas cloud is found to contain a number of dark lines.
(a) Explain why these dark lines occur.
(b) Some electron energy levels in a hydrogen atom are illustrated in
Fig. 8.2.
One dark line is observed at a
wavelength of 435 nm.
(i) Calculate the energy, in eV, of
a photon of light of wavelength 435 nm.
(ii) On Fig. 8.2, draw an arrow to
indicate the energy change that gives rise to this dark line.
Reference: Past Exam Paper – November 2014 Paper 41 & 42
Q8
Solution 1051:
(a) The photon is ‘absorbed’ by the electrons. The photon has energy
equal to the difference in energy of two energy levels. The electron de-excites
emitting photon (of same energy) in any direction.
(b)
(i)
Energy E = hc / λ
E = (6.63×10–34 × 3×108)
/ (435×10–9)
E = 4.57 × 10–19 J
{To convert energy in J to
eV, we divide by the charge of an electron.}
E = (4.57 × 10–19) / (1.6
× 10–19) (eV)
E = 2.86 eV
(ii) arrow pointing in either
direction between –3.41 eV and –0.55 eV
{We need to consider 2
energy levels that have a difference in energy of 2.86eV. These are the energy levels
–3.41 eV and –0.55 eV.}
