Physics91 Quiz
Physics91 Quiz
Physics91 Quiz
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.
light
anode
variable
voltage
source caesium metal cathode
10.2 Calculate:
10.3 Give ONE application, other than a photovoltaic cell, which makes use of the
particle nature of light. (1)
[13]
TOTAL: 150
END
2
PHYSICAL SCIENCES: PHYSICS 18
(Paper 1) 10841/22
MOTION/BEWEGING
Δx v i Δt 21 at 2 or/of
v f vi a t
Δy v i Δt 21 at 2
v vf v vf
v f v i 2ax or/of v f v i 2ay
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
FG or/of FG gG or/of gG
d2 r 2
d2 r2
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
emf ( ε ) = I(R + r)
V
R
I emk ( ε ) = I(R + r)
R s R1 R 2 ...
1 1 1 q It
...
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
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)
= 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
TOTAL/TOTAAL: 150
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