STPM Phys P2 Skema 2011

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JAWATANKUASA PANEL TRIAL EXAM PHYSICS STPM NEGERI SEMBILAN DARUL KHUSUS

MA Rl=-RT 3 ; M=NAm

-H

2

A

(1)

2

So the average kinetic energy for a molecule of the gas is

1 , 3 R -m < c- >= - - T 2 2 NA

=

3 -kT

R

where k=-

2

(c) Total kinetic energy for one mole ofmonoatomic gas

···················(I)

NA

3 R =--T 2 NA

... (I)

=

~ (6.02xl 023 )(1.38xl 0'23 )( 400K)

=

4.99

2

X

I 03 J

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(I)

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CONFIDENTIAL* (d) Number of mocules

Molecular speed

Shape of the graphs (No simetri Curve) Labeling axes and T 1 & T2 Labeling v,, Vm & for both graphs

( 1) ( 1) (!)

v

(e)

The area under graph T 1 and T2 are the same because the number of molecules i remams same. . ................................................................................. (1)

v

The value ofV,. V m and are higher for higher temperature because the kinetic energy of the gas molecule will increase as the temperature increase............................................. (1)

2

12 (a) Two thin conducting plates have the same area of 0.50 m They are placed parallel to each other and 20 mm apart. One plate is maintained at +75 V while the other at- 75 V by a d.c. supply. (i) Detem1ine the amount of charge stored on each plate. [3 marks] (ii) Calculate the energy per unit volume stored in the electric field between the plates. [2 marks] (b) A parallel-plate capacitor of capacitance 3.0 fiF and a resistor of large resistance 1.0 MO are co1111ected in series with a 12 V battery of negligible intemal resistance.

E

s

c (i)

(ii) (iii) (iv)

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R

Sketch graphs to show the time variation of the voltage across the capacitor and the current through it when switch Sis closed. Label clearly the maximum voltage and current on the respective graphs. [4 marks] What is the time constant of the circuit? [1 mark] Calculate the time taken for the capacitor to be 50% charged. [2 marks] With the switch S closed, the plates of the capacitor are pulled slightly apart. Describe the possible changes in the potential difference and the energy store in the capacitor. [3 marks]

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Marking scheme :

I (a)

'"'oA

C=

(i)

=

B.SSx10-!.~xO.SO

2.21 x10- 10

2nx1o-a

d

.................. (!

)

Q=CV =2.21 x 10. 10 x [75-(-75)] ............................... (1) 8 =3.32xi0 C ............................. (l)

(ii) Energy per unit volume=

(112)QV Ad 3.32. · x10- 8 x150

= =

(b)

2.y().5x0.02()

2.49 x I 0"Jm· 3

.......................... (1) ................... (I)

I IA

VN

(i)

12

------------------

1.2xl o·'

(2)

0

tis

(2)

0'-----------+tis (ii)

1

= CR = 3.0 s

....................... (I) -I

(iii) 1

-=1-e 2

J>.o

.................. (!) I

e;;:c; = 2 ...... . .

(I)

(iv) The capacitance of the capacitor decreases because d increases but the potential difference remain constant

. '' ..................................(1)

Therefore the charges on the capacitor decreases

...................... (l)

The energy stored decreases .......................................... (I)

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(a) The process X-rays production is the inverse of that of the photoelectric effect. Explain this statement.

[2 marks] (b)

The graph shows the X -rays spectrum produce by a X -ray tube.

What is the effect of increasing the anode voltage to the (i) characteristic X-rays and continuous X-rays spectra? [2 marks] (ii) What is the effect of using a metal target of higher atomic number? [2 marks] (iii) What is the effect of using a metal target of higher atomic [2 marks] number.

Relative intensity

I

(c)

An X-ray beam is incident on a solid sample. State two conditions that are necessary to observe X-ray diffraction peak from the sample.

I

Wavelength, t..

[2 marks]

(d) A beam ofX-rays of wavelength 0.154 nm is incident on a crystal ofinterplanar spacing 0.288 nm. Calculate (i) the maximum order of diffracted X-rays beams that can be detected. (ii) the Bragg's angle for the highest order of diffraction. [5 marks] Marking scheme : (a) In the photoelectric effect, electrons are emitted from the metal surface when it is illuminated by light, that is photon .............................. (I) In an X-rays tube, electromagnetic radiation, that is photons, are emitted from the metal surface when the metal is hit by electrons. .. ........................ (1) (b) (i) There is a threshold value for anode voltage before X-rays can be generated. Variation in the anode voltage does not alter the energy of the characteristic lines. Increasing the anode voltage will shift the X-ray minimum wavelength towards a lower value, widen the X-rays energy range, and cause overall increase in the Xrav intensity. .. ................................ Any 2 OR Amin decreases, characteristic peaks remain at the same wavelengths but overall increase in the X-ray intensity.

Intensity

0 ;\min

lntensi

Wavelength

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(b) (ii) Increasing the current in the filament will increase overall X-rays intensitv. This is because higher anode current means more electrons reach the target metal result in more X-rays photons. The positions of the characteristics peaks and the cut-offwavelencth are not altered. .. ....... Any 2 OR

Intensity

The positions of the characteristics peaks and A,nm remains unchanged, but intensity increases for all wavelengths.

length

0 -'rnm

Wavelength

(b)(iii) Different metal targets will produce different characteristic X-ray energies because each metal target has . own umgue . 1 . . 1·mes. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (7) 1ts C1aractenstic Intensity

OR

!""'"remains unchanged, but characteristic peaks shift to shorter wavelengths. .......................... (J) (c) The solid must be crystalline The wavelength of the X-ray is comparable to the crystal spacing ..... (1)

(i) (ii)

Wavelength

Maximum order= 3 ......... (1) From Bragg's equation, 2dsin8 = nl,

11

=3

(1)

.

3(0.154) s,ne = 2(0. 288)

e=

53.3° (JJ

14 (a) Define the binding energy per nucleon of a nucleus and mass defect of a'reaction. (b) Calculate the binding energy per nucleon for (i) Uranium-23 8 nucleus, (ii) Carbon-12 nucleus.

[2 marks] [3 marks] [3 marks]

Why is the Uranium-238 nucleus Jess stable than Carbon-12 nucleus?

[!mark]

(Given mass of neutron, mn=l.00867 u ; mass of proton, m,=J.00782 u ; atomic mass of Carbon-12, Mc=l2.00000 u and atomic mass ofUranium-238, Mu=238.05079 u)

(c) An element of unknown mass is mixed with path of the ions) of the element and ions of

'zC in a mass spectrometer. The radii of curvature of (the

'tc are 26.2 em and 22.4 em respectively. What is the element?

State any assumption made.

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[6 marks]

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Marking scheme : l (a) Binding energy per nucleon of a nucleus is the energy required per nucleon to completely separate the . ................. : ... (1). nucleons of the nucleus. Mass defect is the difference between the total mass of the nucleons and the mass of the nucleus ...... (!) (b) (i) For Uranium-238 1.00782 U + 146 X 1.00867 U -238.05079 U = 1.93447 u ............................. (l) 1.9:>447 x931.5 Mev 238 Binding energy per nucleon= A flm = 92

X

= 7.57 MeV

························ (1)

···························· (1)

(ii) For Carbon-12 flm = (&mP + 6n~)- Me = 6(1.00782 u + 1.00867 u)- 12.00000 u = 0.09894 u .................. (1)

L!..·mcz Binding energy per nucleon =

A

O.G9394 x 931..5

12

...................... (1)

............................. (I) = 7.68 MeV Since the binding energy per nucleon ofuraninm-23 8 is less than the binding energy per nucleon of carbon-12. Therefore uranium-238 is less stable than carbon-12 .................. (1) mVZ

qvB=r

(c)

················· (I)

m= (~)r 1

Assuming that the ions of the element have the same charge as the ions of therefore

m ocr A ocr A

2;6.2

12

22.4

~c,

.............. (1)

.................. (I)

---

(1) .................... (I) ................... (I)

......................

A = 14 The element is nitrogen

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