trialstpm2013YUHUApaper3

SMJK Yu Hua, Kajang, Selangor
STPM Trial Examination
September 2013
Physics Paper 3

Form : 6AS1 Prepared by : Pn Rosmaya Mokhtar
Date :23rd September 2013 (Monday) Verified by : En. Chiew kiang Keng
Time : 11:35 – 1:05 am

Number of printed pages : 4

Name:……………………………………………….

Section A. (15 marks)

Answer all questions.

1. The physical process not observable for sound waves is

A. Interference C. Reflection

B. Polarization D. Refraction

2. In which group below do the three quantities remain constant when a particle moves in simple harmonic

motion?

A Force Acceleration Total energy

B Total energy Force Amplitude

C Amplitude Total energy Angular frequency

D Angular frequency Amplitude Acceleration

3. A sound wave with frequency 200 Hz moves in air with speed 320 ms-1. What is the phase difference

between two points on the wave which are separated by 0.2 m?

A.  rad B.  rad C. 2 rad D 4 rad

4 25 5

4. A cyclist approaches a songbird on a tree at a speed of 10 ms-1. If the bird sings at a frequency of 420 Hz,

what is the frequency heard by the cyclist?

[speed of sound in air = 343 ms-1]

A. 396 Hz B. 408 Hz C. 432 Hz D. 445 Hz

5. A beat frequency of 9 Hz is detected when a musical note is sounded simultaneously with a tuning fork of

frequency 440 Hz. If the same musical note is sounded with another tuning fork of 442 Hz, the beat

frequency that is detected is more than 9 Hz. The frequency of the musical note is

A. 431 Hz B. 433 Hz D. 445 Hz D. 449 Hz

6. The diagram below shows a bulb placed in front of an arrangement of convex lens and concave lens with
focal lengths of 20 cm and 10 cm respectively. The lens are arranged 20 cm apart along the same axis.


Bulb

20 cm

What is the distance of the bulb from the convex lens so that the light rays emerging from the concave

lens are parallel?

A. 10 cm B. 12 cm C. 20 cm D. 60 cm

1

7. If the distance between a real object and a real image produced by a converging lens is minimum and

equal to d, the objects distance from the lens is

A. 3 d B. d C. d D. 2 d

16 3 23

8. The figure shows monochromatic light of wavelength  which illuminates at nearly normal incidence a
thin soap film of thickness t and refractive index n.

air

soap t

air

The optical path difference between the reflected rays from the upper and lower surfaces of the soap film

is

A. 2t B. 2nt C. 2t +  D. 2nt + 

n n2 2

9. When a ray of monochromatic light is incident normally on a single slit, a diffraction pattern consisting of

a central maximum and several subsidiary maxima is observed. When the whole system is immersed

completely in a transparent liquid, which of the following statements describes the change to the original

interference pattern?

A. The width of the central maximum decreases

B. Only the central maximum is observed

C. No diffraction pattern can be seen

D. The diffraction pattern is displaced from the original place.

10. The diagram below shows two energy levels of an atom. What is the wavelength of the photon produced
when an energy transition happens between the two energy levels?

– 1.51 eV

photon

– 13.6 eV

A. 8.23 x 10-4 m B. 9.14 x 10-8 m C. 1.03 x 10-7 m D. 1.65 x 10-7 m

2

11. The diagram below shows the variation of intensity with the wavelength of the X-rays from an X-ray
tube.

intensity

0 min wavelength

Which of the following describes correctly the change, if any, in min and the intensity of the continuous
spectrum in the diagram above if the potential difference of the X-ray tube is increased?

min Intensity of the continuous spectrum
A decreases
B decreases decreases
C no change increases
D increases increases
decreases

12. The diagram below shows a diffraction image of the second order obtained from a beam of diffracted
X-ray.

Incident X-ray beam Diffracted X-ray beam

128o

Atomic plane
•••

d

Atomic plane
•••

If the wavelength of X-ray is 1.10 x 10-10 m, the separation d between the atomic planes is

A. 1.22 x 10-10 m B. 1.25 x 10-10 m C. 2.44 x 10-10 m D. 2.51 x 10-10 m

13. According to definition, binding energy per nucleon can be used as a measurement for the stability of a
nucleus. This energy will
A. increase if the value of ratio of neutron/proton of the nucleus increases
B. increase uniformly for the whole periodic table
C. be maximum for nucleons in the middle part of the periodic table
D. decrease to zero for radioactive nucleons that are heavy

14. Isotope 234Th with a half-life of 24 days decays to form 234Pa. How long does it take for 90% of the

90 91

isotope sample to change into 234Pa ?

91

A. 5 days B. 13 days C. 22 days D. 80 days

15. In the choice of a radioisotope as a tracer in medicine, agriculture and industry, which of the following is

the least considered?

A. Nucleus C. Intensity of emitted radiation

B Charge of emitted particle. D. Half-life of radioisotope

3

Section B. (15 marks) [1 mark]
Answer all questions. [2 marks]
16. The equation of a progressive wave in a taut string is [2 marks]
[3 marks]
y = 0.02 sin (4t – 0.25x)
where x and y are in units of meters, while t is in units of seconds.
a) What do x and y represent?
b) What is the velocity of the progressive wave?
c) What is the displacement of a particle if x is 0.15 m and t is 0.30 s ?
d) What is the speed at the instant of situation (c)?

17. a) What is nuclear fusion? [1 mark]

b) Describe the thermonuclear fusion in the Sun. [3 marks]

c) The intensity of radiation of the Sun on the Earth is 1340 W m-2. The mean distance of the Earth from

the Sun 1.50 x 1011 m. Calculate the radiation energy of the Sun in a year. (1 year = 3.15 x 107 s)

[3 marks]

Section C. (30 marks)

Answer any two questions.

18. a) State the definition of interference [1 marks]

b) i) Explain what is meant by coherence between two sources of light.. [2 marks]

ii) State whether or not two sources emitting light of different wavelengths are coherent.

[2 marks]

iii) State one more condition other than coherence, for obtaining clear interference patterns

between two rays of light. Explain why the interference pattern cannot be observed if that

condition is not satisfied. [2 marks]

c) In a Young’s double slit experiment, the separation between the two slits is 0.05 cm, the distance

between the slits and the screen is 200 cm. When red light is used, the distance between the centre of

the interference pattern and the first fringe is 0.13 cm.

i) Calculate the wavelength of the red light used. [2 marks]

ii) Calculate the distance between the centre of the interference pattern with the third dark

fringe. [2 marks]

iii) If the space between Young’s slits and the screen is filled with water, explain what will

happen to the interference pattern on the screen. [4 marks]

4

19. a) What is photoelectric emission? [1 marks]

b) Einstein’s photoelectric equation is given as follows :

hf = ½ mevm2 + W [3 marks]
i) State the meaning of the terms hf, ½ mevm2 and W in the equation above.

ii) With the help of an equation containing W, state the condition for the frequency of the

incident radiation that will allow photoelectric emission. [2 marks]

iii) Explain why photoelectrons are emitted with varying velocities even though radiation of

constant frequency is being used. [3 marks]

c) In a photoelectric experiment, the stopping potential difference for incident radiation of wavelength

300 nm and 400 nm are respectively 1.8 V and 0.8 V.
iv) Explain what is meant by the ‘stopping potential difference’ and hence, calculate the

maximum velocity of the electrons emitted by an incident radiation of wavelength 300 nm.

[3 marks]

v) Estimate the value of the Planck constant from the information above. [3 marks]

20. a) i) Clarify what is meant by radioactive decay and the activity of a radioactive source. [2 marks]
b)
ii) Give the relationship between the activity of a radioactive source and the number of

radioactive nucleus in the source. Hence, define decay constant. [2 marks]

iii) Define the half-life of a radioactive source and show how the half-life is associated with the

decay constant. [4 marks]

A radioactive point source of decay constant 6.0 x 10-11 s-1 decays by emitting -particles with the

same probability in all directions. The number of -particles detected in an area of 2.0 cm2 at a

distance of 10 m from the radioactive source is 1000 per second. By assuming that each of the

nucleus in the radioactive point source decays to form a stable nucleus, determine

i) the number of radioactive nucleus in the point source. [3 marks]

ii) the fraction of the radioactive source that remains after 50 years. [2 marks]

iii) the activity of the radioactive source after 50 years. [2 marks]

5

Mark scheme for paper 3 Physics (Trial STPM 2013)
Section A.
1. B.
2. D
3. A
4. C
5. A
6. D
7. C
8. D
9. A
10. C
11. B
12. D
13. A
14. D
15. B

Section B.
16. a) x is a variable i.e distance of a particle from the source of the wave

y is another variable i.e vertical displacement of particle at time t.
b) y = A sin [t - 2x/]

y = 0.02 sin (4t – 0.25x)
=4
2f = 4

f = 2/
0.25 = 2/

 = 8.0
v = f

=2/ x 8
= 16 ms-1
c) y = 0.02 sin (4t – 0.25x)
=0.02 sin (4(0.3) – 0.25(0.15))
= 0.0187 m
c) y = 0.02 sin (4t – 0.25(0.15))
= 0.02 sin (4.0t – 0.0375)
v = dy/dt
v = 0.08 cos (4.0(0.3) – 0.0375)
= 0.08 cos 1.196
= 0.025 ms-1

17. a) The joining together of the small nuclei to form a larger nucleus
b) Due to the very high temperature (> 107K) in the Sun, hydrogen atoms travel at high speed. With
sufficiently high kinetic energy to overcome the repulsive electrostatic force, fusion between hydrogen
atoms occur. During each fusion, the energy equivalent of the mass defect is radiated.
c) I = P/4R2
P = 4R2I
Energy radiated in a year = Pt
= 4R2It
= 4(1.50 x 1011)2(1340)(3.15 x 107)
= 1.19 x 1034 J

Section C.
18. a) The effect produced due to the superposition of waves from two coherent sources moving through

6

the same region

b) i) If the light waves emitted by the two sources have the same amplitude, same frequency and

the phase difference between them is zero or is constant with time, then the two sources are

said to be coherent

ii) If two series of waves have different wavelengths, then the frequency of the series of waves

are not the same. The phase difference between them will not be zero or constant but will

change with time. By definition, the two series of waves are not coherent.

iii) Another condition is that the separation of the two sources must be near enough to satisfy the

expression of the width of the fringes x, given by

x = D/a

where D = distance of screen from the slit

 = wavelength

a = separation of the slits

c) i) x/D = /a

0.13/200 = /0.05

 = 3.25 x 10-5 cm

= 3.25 x 10-7 m

ii) x/200 = 2.5 (3.25 x 10-5)/0.05

x = 0.325 cm

iii) It is known that the refractive index of water is higher that that of air. Water has a higher

density than air. The frequency of light in air is the same as that in water. The wavelength is

however shorter in water than that in air. Based on this information, we still observe same

interference pattern, except that the fringes will be decreased. This is because x = D/a.

x = separation of two adjacent bright fringes

D = distance of screen from the slits

a = separation of the two slits

D and a are constants. If  decreases, x will also decrease.

19. a) Photoelectric emission is the emission of electrons from the surface of a metal when electromagnetic

radiation of a certain frequency is incident upon the metal.

b) i) hf = energy for a ‘packet’ of incident radiation or of a photon

½ mvm2 = maximum kinetic energy of the emitted electrons

W = work function ie the work needed to be done to eject an electron from the outer surface

receiving the radiation. This is the energy needed to overcome the potential barrier.

ii) E = hf = ½ mvm2 + W

from the formula, we can conclude that no electron can be emitted if hf is less than W.

Therefore there exists a threshold frequency.

Therefore the frequency of the incident radiation must exceed a minimum frequency which is

the threshold frequency before photoelectric emission can occur. This minimum frequency

supplies energy hfo for the electron to be extracted from the outer surface.

vi) W = hfo
The photoelectric equation can be stated as h(f – fo) = ½ mvm2

So, a fix frequency of the incident radiation will cause electrons to be emitted with the same

velocity. However, the energy needed to eject an electron from the metal surface varies. More

energy is needed to remove an electron farther away from the surface. This means that

electrons nearer to the surface will be ejected with higher velocity. We can conclude that

different electrons from different places in the metal will be ejected with different velocity.

c) i) The stopping potential is the minimum potential difference which must be maintained across

the cathode and the surface to prevent any electron from reaching the cathode.
eV = ½ mvm2

1.6 x 10-19 x 1.8 = ½ x 9.1 x 10-31 vm2
vm = 7.96 x 105 ms-1

7

iv) hf = eV + W
hf1 = eV1 + W
hf2 = eV2 + W
h(f1 – f2) = e(V1 – V2)
h = e (V1 – V2)
f1 – f2
= e (V1 – V2)
c/1 – c/2
= 1.6 x 10-19 (1.8 – 0.8)
3 x 108(1/300 x 10-8 – 1/400 x 10-9)
= 6.4 x 10-24 Js

20. a) i) Radioactive decay is the process where an unstable nucleus becomes another stable nucleus by
emitting -radiation, -radiation or -radiation.
Activity of a radioactive source is the rate of decay of the radioactive source, dN/dt or the number
of radioactive atoms of the source which decays in a unit time period.

ii) dN/dt = - N
N = number of atoms which has not decayed
Decay constant is the rate of decay of an undecayed atom

iii) Half-life is the time it takes for a radioactive sample to decay to half its original number of
radioactive nuclei or the time taken for half the atoms of the radioactive sample to decay.
N = Noe-t
At time T ½ , N = No/2
e-t = ½
-T ½ = - ln 2
T ½ = ln 2/

b) i) Number of -particles at distance of 10 cm is 1000 per 2.0 cm
Total number of -particles emitted =1000/2 (4 x 102)
= 6.28 x 105 s-1
N = dN/dt \ 
= 6.28 x 105
6.0 x 1011
= 1.05 x 1016 atoms

ii) N = Noe-t
t = 6.0 x 1011 x 50 x 365 x 24 x 60 x 60
= 9.46 x 10-2
N/No = e9.46 10-2
= 0.9097

iii) dN/dt = N
= 5.731 x 105 s-1

8