Mark Scheme Trial PHYSICS 2013 Sem 3

MARK SCHEME SECTION A NO ANSWER EXPLAINATION 1. B 2. C 3. D F = 0.5, λ= 1/0.1 V = fλ = 0.5(1/0.1)= 5.0 ms-1 4. A I = ( ) = 0.32 Wm-2 5. B L = = = 22.67 x 10-3 = 0.02267 m = 2.3 cm 6. B F1 = ( ) = 206.25 Hz F2 = ( ) Beat frequency = (212.5 -206.25 ) Hz = 6.25 Hz 7. C 8. A but u’ = v = v’ = f’ = -20 cm

Distance of emerging ray from the first lens = 25 – 20 = 5 cm

10. C = 2.06 x 10-7 m From c = f F = 11. D 12. B E hc hc hf E E E _{f i}        





13. A

The K characteristic line is produced by electronic transition in the target atom. The intensity is determined by the number of electrons hitting the target and the number of collisions determines the number of electronic transition.

14. B E is the energy equivalent of



m

.

15. B

hours

e

e

N

N

t

10

0693

.

0

2

ln

2

ln

life

half

0693

.

0

20

4

ln

40

10

(3010) 0

















  







 

SECTION B

16a A note of higher frequency than the fundamental note. 2

b-i

For the first resonance:



l₁c



4 

v

4 f (i)

For the second resonance:

 l₂c3



4  3v 4 f (ii)  (ii) (i) l₂l₁ v 2 f speed of sound, v2 f (l₂l₁)2(250 )(0.960.30) 330 ms-1 1 1 1 1 b-ii

Substitute the value of v into (i)



0.30c 330 4(250)

 the end correction, c0.03 m

1 1 17 a-i









0 0 49 00 . 1 30 510 . 1   red red i Sini Sin 1 1 a-ii









0 0 5 . 49 00 . 1 30 521 . 1   blue blue i Sini Sin 1 1 a-iii Angle between the red and the blue refracted rays

0 0 0 5 . 0 49 5 . 49    1 b





 









 





0 7 5 1 7 5 1 1 1 0 . 4 10 9 . 4 1 10 0 . 4 10 66 . 1 1 10 0 . 4 1 1                                    Sin Sin n d Sin n d Sin blue red 1 1 SECTION C 18 (a)

the acceleration of an object undergoing oscillation

i. is directly proportional to the displacement of the object from the equilibrium position

ii. always points toward the equilibrium position

1 1

- the string has to be inelastic - (c) i. T = 2



g L = 2.00 s ii. Ki + Ui = Kf +Uf 0 +mgh = 2 1 mv2 + 0 v2 = 2gh = 2(9.81)(1 – cos 100) = 0.55 ms-1 1 1 1 1 1 (d) i. k= mw2 = 0.5(2



f)2 = 33.4 Nm-1 1 1 ii. E = 2 1 kA2 0.5 = ½(33.4)A2 A=0.17 m 1 1 19 (a)(i)

The quantum of electromagnetic radiation , having an energy related to the frequency f, of the light by E = hf where E is the photon energy and h is Planck’s

constant. 1

(a)(ii)

Quantum theory include:

1. the effect of frequency on the photoelectric effect 2. the existence of threshold frequency

3. the maximum kinetic energy of photoelectrons does not depend on the incident light intensity

4. the instantaneous emission of the photoelectrons

1 1 1 1 (b)(i) 

Maximum kinetic energy, 1 2mvm 2 _ hc



W  (6.631034₎ 3.00108 4.13107            



20(1.61019₎



1.621019 _J 1 1 1

(b)(ii)



1

2

mv

m 2

_

_1.62

_

₁₀

19

_J

mv

_m2

_

_2(1.6

_

₁₀

19

_)m

mv

_m



2(1.6



10

19

_)(9.11

_

₁₀

31

₎

de Broglie wavelength of electron,





h

mv

_m



6.63



10

34

2(1.6



10

19

_)(9.11

_

₁₀

31

₎



1.22



10

9

_m

1 1 1 (b)(iii)

If n is the number of electrons per second falling on the cesium surface, hence

 nhc



IA nIA



hc  (5)(25104_)(4.13107₎ (6.631034_)(3108₎

The number of electrons per second extracted 60%(5)(25104)(4.13107) (6.631034_)(3108₎ 1.561016 1 1 1 1 20 (a) (i)

An α-particle is a helium nucleus which consists of 2 protons and 2 neutrons. When an atom X emits an α-particle, the new atom produced will have its mass number reduced by 4 and its atomic number reduced by 2.

X He AZ Y A Z 4 2 4 2     2

(ii) A β-particle is high velocity electron. When an atom emits a β-particle , a neutron in its nucleus has been converted to a proton and a β-particle.



0 1 1 1 1 0n p 1 1

(iii) γ-particle is an electromagnetic radiation which has energy but no mass. It is emitted when a radioactive nucleus converts its mass defect into energy in order to become another more stable nucleus.

1 1 (b)(i) Radioactive decay refers to the random and spontaneous process whereby a less stable

nucleus changes to a more stable nuclide with the emission of radioactive radiation. Activity is the number of radioactive atoms decay in unit time.

3

(ii)

The activity dt dN

N dt dN N dt dN 1      





Hence, the decay constant λ can be defined as the ratio of the activity to the number of radioactive nuclei in the source before decay.

(iii) Half life is the time taken for a radioactive nuclide sample to decay until half of its initial number of nuclei are left behind.

From dt N

dN

_

 

If N0 is the number of nuclei at time t= 0. and N is the number of nuclei at time t,















  693 . 0 2 ln 2 ln 2 , 2 , , ln 2 1 2 1 2 / 1 0 0 0 0 0                





T T e half time T t N N if e N N t N dt N dN t t N N t N N 4