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Practice Quiz |Physical Science:91|Class

PHYSICAL SCIENCES: PHYSICS


:12th
17
(Paper 1) 10841/22
Questions and Answers
QUESTION 10 (Start on a new page.)

A photoelectric cell is connected in a circuit. The lowest frequency of light that will emit
14
electrons from its caesium surface is 5,1 x 10 Hz.

Violet light of wavelength 400 nm is incident on the caesium surface.

light
anode

variable
voltage
source caesium metal cathode

10.1 Define threshold frequency. (2)

10.2 Calculate:

10.2.1 The work function of caesium (3)

10.2.2 The amount of energy carried by the incident photons of


violet light (3)

10.2.3 The maximum kinetic energy of the photoelectrons emitted


from the caesium surface when violet light shines on it (4)

10.3 Give ONE application, other than a photovoltaic cell, which makes use of the
particle nature of light. (1)
[13]

TOTAL: 150

END
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PHYSICAL SCIENCES: PHYSICS 18
(Paper 1) 10841/22

DATA FOR PHYSICAL SCIENCES GRADE 12


PAPER 1 (PHYSICS)

GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 12


VRAESTEL 1 (FISIKA)

TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES

NAME/NAAM SYMBOL/SIMBOOL VALUE/WAARDE


Acceleration due to gravity
g 9,8 m·s-2
Swaartekragversnelling
Universal gravitational constant
G 6,67 x 10-11 N·m2·kg-2
Universele gravitasiekonstant
Radius of the Earth
RE 6,38 x 106 m
Radius van die Aarde
Mass of the Earth
ME 5,98 x 1024 kg
Massa van die Aarde
Speed of light in a vacuum
c 3,0 x 108 m·s-1
Spoed van lig in 'n vakuum
Planck's constant
h 6,63 x 10-34 J·s
Planck se konstante
Coulomb's constant
k 9,0 x 109 N·m2·C-2
Coulomb se konstante
Charge on electron
e -1,6 x 10-19 C
Lading op elektron
Electron mass
me 9,11 x 10-31 kg
Elektronmassa
3
PHYSICAL SCIENCES: PHYSICS 19
(Paper 1) 10841/22

TABLE 2: FORMULAE/TABEL 2: FORMULES

MOTION/BEWEGING

Δx  v i Δt  21 at 2 or/of
v f  vi  a t
Δy  v i Δt  21 at 2
 v  vf   v  vf 
v f  v i  2ax or/of v f  v i  2ay
2 2 2 2
Δx   i  Δt or/of Δy   i  Δt
 2   2 

FORCE/KRAG

Fnet  ma p  mv
fsmax   sN fk  kN
Fnet t  p
w  mg
p  mv f  mv i
m 1m2 m 1m2 M M
FG or/of FG gG or/of gG
d2 r 2
d2 r2

WORK, ENERGY AND POWER/ARBEID, ENERGIE EN DRYWING

W  Fx cos  U  mgh or/of EP  mgh


Wnet  K or/of Wnet  Ek
1 1
K  mv 2 or/of Ek  mv 2
2 2
K  K f  K i or/of Ek  Ekf  Eki
W
Wnc  K  U or/of Wnc  E k  E p P
t
Pav e  Fv av e /
Pgem  Fv gem

WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG

1
v f T
f
v  vL v  vL c
fL  fs fL  fb E  hf or/of E h
v  vs v  vb 
E  Wo  Ek(max) or/of E  Wo  K max where/waar
1 1
E  hf and/en W0  hf 0 and/en Ek(max)  mv max
2
or/of K max  mv max
2

2 2
4
PHYSICAL SCIENCES: PHYSICS 20
(Paper 1) 10841/22

ELECTROSTATICS/ELEKTROSTATIKA

kQ 1Q 2 kQ
F E
r2 r2
W F
V E
q q
Q Q
n or/of n
e qe

ELECTRIC CIRCUITS/ELEKTRIESE STROOMBANE

emf ( ε ) = I(R + r)
V
R
I emk ( ε ) = I(R + r)
R s  R1  R 2  ...
1 1 1 q  It
   ...
R p R1 R 2
W
W = Vq P
Δt
W = VI  t
P = VI
W= I2R  t
P = I2R
V 2 Δt
W= V2
R P
R

ALTERNATING CURRENT/WISSELSTROOM

Pav e  VrmsIrms / Pgemiddeld Vwgk Iwgk


I max I maks
I rms  / Iwgk 
2 2
Pav e  I rms
2
R / Pgemiddeld  Iwgk
2
R
Vmax Vmaks
Vrms  / Vwgk  V2 Vwgk
2
2 2 Pav e  rms / Pgemiddeld 
R R
5
PHYSICAL SCIENCES: PHYSICS 21
(Paper 1) 10841/22

ANSWER SHEET
QUESTION 8.1
NAME OF LEARNER: ________________________
ANSWERS 6 Practice Quiz |Physical Science:91
10841/22
QUESTION/VRAAG 10

10.1 The minimum frequency of light needed to emit electrons from a certain metal
surface 
Die minimum frekwensie van lig benodig om elektrone vry te stel vanuit ʼn
sekere metaal se oppervlakte
OR/OF
A certain minimum frequency of incident radiation which will cause photoelectrons
to be emitted from the surface of a metal✓✓
ʼn Sekere minimum frekwensie van invallende strale wat sal veroorsaak dat foto-
elektrone uit die oppervlak van ʼn metaal vrygestel word

Marking criteria/Nasienriglyne:
If any of the underlined words/phrases in the correct context is omitted, deduct 1
mark.
If learner explains minimum energy then zero marks are allocated. It has to be
frequency.
Indien enige van die onderstreepte woorde/frases in die korrekte konteks
uitgelaat word, trek 1 punt af.
Indien leerder minimum energie verduidelik, dan geen punte toegeken nie. Dit
moet frekwensie wees. (2)

10.2 10.2.1 𝑊0 = ℎ𝑓0  Marking guidelines/


= 6,63 x 10-34 5,1 x 1014  Nasienriglyne:
= 3,38 x 10-19 J  ✓ Formula/Formule
✓ Substitution/Vervanging
✓ Answer/Antwoord (3)

10.2.2 OPTION/OPSIE 1 Marking guidelines/


𝑐 Nasienriglyne:
E= ℎ   ✓ Formula/Formule
= 6,63 x 10-34 x 3 x 108  ✓ ✓Substitution/Vervanging
400 x 10-9✓ ✓ Answer/Antwoord

= 4,97 x 10-19 J 

20
7
10841/22

OPTION 2: OPSIE 2

𝑐
𝑓 = 
= 3 x 108 ✓
400 x 10-9

= 7,50 x 1014 Hz
E= ℎ𝑓
= 6,63 x 10-34 x 7,5 x 1014 

= 4,97 x 10-19 J  (4)

10.2.3 Positive marking from Marking


Question 10.2.1 and 10.2.2 guidelines/Nasienriglyne:
Positiewe nasien vanaf Vraag ✓ Formula/Formule
10.2.1 en 10.2.2 ✓ ✓Substitution/Vervanging
𝑐
✓ Answer/Antwoord
E= ℎ  = 𝑊0 + 𝐸𝑘max/𝑚𝑎𝑘𝑠 
4,97 x 10-19 ✓ = 3,38 x 10-19 + Notes/Notas:
Ekmax/maks✓
• If max. is omitted from Ek
Ekmax/maks = 1,59 x 10-19 J  formula, marks are lost for the
formula. The rest can however
be marked./
• Indien maks. uitgelaat word
van formule met Ek, word
punte verbeur vir die formule.
Die res kan wel steeds punte
kry. (4)

10.3 Not to be marked. Mark moved to 10.2.2


[13]

TOTAL/TOTAAL: 150

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