0 Mathematics I Heat and Gases 13 Number of mole n, number of molecules N, mass of a gas 6 Acceleration due to gravity g = 9.

81 m s–2
molecule m, molar mass M of the gas:
(direction: vertically downwards )
1 Common SI prefixes: 1 Temperature-dependent T NA = Avogadro number = 6.02  1023 mol–1
Prefix Factor Prefix Factor property x (e.g. length of
TU N = nNA and M = mNA 7 Motion graphs s slope = V
liquid column)
tera- (T) 1012 milli- (m) 10–3
Temperature is TL at lower ( straight line motions only):
giga- (G) 109 micro- (μ) 10–6 fixed point xL, temperature is
T 14 Mean square value of velocities of molecules ̅̅̅
𝑐 2 ( m 2s–2 ) S

TU at upper fixed point xU; and root-mean-square speed crms ( m s–1 ): Slope of s–t graph = velocity t
mega- (M) 106 nano- (n) 10–9 TL
temperature T at x is 3RT Slope of v–t graph = acceleration v
kilo- (k) 103 pico- (p) 10–12 crms = √̅̅̅
𝑐2 = √
x – xL T – TL M
= Area under v–t graph = displacement V
xU – xL TU – TL x
million = 106 billion = 109 xL x xU
15 Equation relating p, V, N, m and c̅2 is
1
t
pV = Nmc̅2
2 Addition of vectors 𝑐⃗ = 𝑎⃗ + 𝑏⃗⃗ 2 Heater of power P is switched on for time t to transfer energy
3 area = S
macroscopic microscopic
Q. 8 An object is thrown upwards with speed u. The motion graphs
𝑏⃗⃗ 16 Average kinetic energy of a gas molecule is
are (upwards as +ve):
Q
P= 1 3RT
t KEaverage = mv2 =
𝑎⃗ 𝑎⃗ 𝑐⃗ 2 2NA s
–1
𝑐⃗ Unit: W s J (W = J s )
Unit: J K
s0
𝑏⃗⃗ 3 1 kilowatt-hour (kW h) = 1000 W × 3600 s = 3.6 MJ
17 Internal energy of n mole of ideal gas at temperature T is
3 Trigonometric ratios: 4 Object of mass m and specific heat capacity c 3 t
absorbs/releases heat Q. Temperature change is T. Internal energy = nRT
a 2
sin  =
c Q = mc∆T Unit: J K v
c
a
b
cos  = Unit: J kg J kg–1 °C–1 °C u slope = –9.81 m s–2
c 
II Force and Motion area = s0
a b 5 Heat capacity C (of object of mass m) and specific heat
tan  = t
b capacity c (of material): 1 Object moves a distance d in time t. 𝑠⃗
y d
4 Slope m of straight line C = mc d
y2 – y1
Average speed = –u
t
m= (x2, y2) Unit: J °C –1
kg J kg °C–1 –1
x2 – x1 Unit: m s–1 s m
= tan 
y2 – y1 6 Object of mass m and specific latent heat l absorbs heat Q a
(x1, y1)  2 Object has a displacement s⃗⃗ in time t.
and changes its state (fusion or vaporization).
x2 – x1 s
⃗⃗
Q = ml Average velocity =
c t
m>0 m=0 m0 x –1 Unit: m s–1 s m t
Unit: J kg J kg
1000 m 1
5 Equation of straight line (slope m and y-intercept c) 7 Kelvin temperature TK and Celsius temperature TC: 3 1 km h–1 = = m s–1 –9.81 m s–2
3600 s 3.6
y = mx + c
TK = Tc + 273
4 Velocity of an object changes from v⃗⃗⃗⃗1 to v⃗⃗⃗⃗2 in time t.
6 Direct proportion and inverse proportion:
K °C
Average acceleration =
v2 − ⃗⃗⃗⃗⃗
⃗⃗⃗⃗⃗ 𝑣1 v2 − ⃗⃗⃗⃗
⃗⃗⃗⃗ v1 9 Newton’s laws of motion (mass m, acceleration a)
a is directly proportional to b a is inversely proportional to 8 Internal energy of object t - First law:
v1
⃗⃗⃗⃗ at rest
1
(a  b) if a
b (___________) if
b = total molecular KE + total molecular PE
Unit: m s–2 s m s–1
v2
⃗⃗⃗⃗ Net force = 0  or
a = kb mass uniform motion (constant velocity)
(where k is a constant) k = no. of molecules  ( average molecular KE +
a= state
b
T
5 An object moves with uniform acceleration a ( m s–2 ) along - Second law:
average molecular PE )
a straight line. It starts with initial velocity u ( m s–1 ) and Magnitude of net force FN = ma
straight line
reaches final velocity v ( m s–1 ) after travelling for time t
a m s–2
a a (For 9-17) A gas of pressure p ( Pa ), volume V ( m 3 ), kelvin ( s ) and displacement s ( m ). Unit: N kg (N = kg m s–2)
temperature T ( K ) has n moles of molecules.
u v - Third law: F12 = –F21
slope = k R = universal gas constant = 8.31 J mol –1 K–1
initial s after 10 Free-body diagram of object resting on rough inclined plane:
b 1
9 Boyle’s law: pV = constant (constant T) position time t
b b normal force
p
pass through origin 10 Pressure law: = constant (constant V)
T Equations of uniformly accelerated motion:

V v = u + at friction
11 Charles’ law: = constant (constant p)
T 1
7 Solve for x in ax2 + bx + c = 0. s = (u + v)t
2
−b ± √b2 − 4ac 12 General gas law: pV = nRT
x= (quadratic formula) 1 weight
2a s = ut + at2
2
8 Sphere of radius r : v2 = u2 + 2as
4
Area A = 4πr2 Volume V = πr3
3 **Take XXX as positive.

1
11 The weight W of an object of mass m is 21 An object is moving with velocity v under action of force F. (For 28-34) Uniform circular motion: object moving in circular path 35 Newton’s law of universal gravitation:
Power P due to F is Gravitational force F between two objects (of masses m 1 and
W = mg at a constant speed.
P = Fv m 2) separated by distance r is:
Unit: N kg m s–2 m2
m s–1 Gm1 m2
Unit: W N 28 Angular displacement F=
s r2 r
m1
s θ F
12 Resolving components (object on smooth inclined plane) 22 An object of mass m is moving with velocity v. = r
Unit: N kg m kg
F
r
Momentum p = mv Unit: rad where G = 6.67  10–11 N m 2 kg–2
Along the plane Along vertical and horizontal
Unit: kg m s–1 kg m s–1
One complete circle s = 2πr ,  = 2π 36 An object of mass m experiences gravitational force F at a
N N N cos θ
position. Gravitational field strength g at the position is
23 Types of collision:
29 The object sweeps angle θ in time t. F
elastic: KEinitial = KEfinal g=
W sin θ θ m
𝜃
N sin θ Angular velocity  = Unit: N kg–1 kg N
θ inelastic: KEinitial > KEfinal t
W cos θ Unit: rad s –1
s rad
completely inelastic: objects stick together 37 Gravitational field strength g at distance r from the centre of
W W
planet of mass M:
24 Law of conservation of momentum: 30 Angular velocity , linear velocity v and radius of path r:
13 Force F acts at perpendicular distance d from point O. GM r
v = r g=
Moment  of F about O is F before collision after collision
r2
M
τ = Fd Unit: N kg–1 m kg
O u1 u2 v1 v2 m s–1 m rad s–1
Unit: N m N m 38 Velocity v of an object in circular orbit (radius r) around a
d m1 m2 m1 m2 31 Period T (time to move one cycle):
body of mass M by gravity:
2𝜋r 2𝜋
14 Equilibrium: T= =
m 1u1 + m 2u2 = m 1v1 + m 2v2 v 𝜔 GM
v=√
Net force = 0 and s r
**Take XXX as positive. m s–1
Unit: m kg
net moment = 0 Assumption: no external net force acting on the system 32 Centripetal acceleration a (towards centre )
(clockwise moment = anticlockwise moment)
v2
25 Momentum of an object changes from mu to mv in time t. a= = r 2
15 An object moves displacement s under action of force F. r
Average net force F acting on object is
Angle between F and s is . mv−mu
Unit: m s–2
F F=
Work done W by force is t
33 Centripetal force F ( resultant force towards centre
W = Fs cos θ θ s 26 Area under F-t graph = change in momentum that causes uniform circular motion) for object of mass m:
Unit: (J = N m) mv2
J N m
27 An object is projected with speed u at an angle  to the F= = mr 2
r
horizontal. kg m m s–1
16 An object of mass m is moving with velocity v. Unit: N rad s–1
ux = u cos θ uy = u sin θ
1
KE = mv2 Horizontal motion: 34 Work done by centripetal force = 0
2
J kg ms –1
sx = (u cos θ)t vx = u cos θ ∵ perpendicular to motion

17 An object of mass m moves through (vertical) height h. Vertical motion:

PE = mgh ay = –g
h 1 1
J kg m sy = uyt + ayt2 = (u sin θ)t – gt2
2 2

vy = uy + ayt =u sin θ – gt
18 Conservation of energy (frictionless motion):
vy2 = uy2 + 2aysy = (u sin θ)2 – 2gsy
KE + PE = constant
(KE + PE) = 0 +ve
vy = 0
Loss in PE = gain in KE t
u +ve
v
19 Conservation of energy (moving through a distance s with
friction or air resistance f): sy sy = 0
t=0 θ
Loss in sum of KE and PE = fs

20 Power P of an object when it transfers energy E in time t by sx

doing work is
E W
P= =
t t

Unit: W s J (W = J s–1)

2
IIIa Wave Motion: Light (Optics) 9 Focal plane: 11 Position and nature of images for concave lenses
Convex: Concave:
1 Laws of reflection normal
incident reflected
(1) Incident ray, reflected ray ray ray
and normal on same plane i r
(2) r = i
F F'

2 Laws of refraction normal

same side as object;
incident Rays do NOT stop here!
(1) Incident ray, refracted ray ray within or at F';
and normal on same plane i
10 Position and nature of images for convex lenses virtual, erect, diminished
sin i
(2) = constant (Snell’s law) Image Nature of
sin r refracted Object position
r ray position image
12 Object height and image height are ho and hi. Object distance
At infinity
and image distance are u and v. Magnification m of image:
3 A light ray hits medium from vacuum at angle of incidence v. ℎ𝑖 v
real, m= =
Angle of refraction is m. Refractive index n of medium: F 2F on focal ℎ𝑜 u
inverted,
sin 𝜃v
plane
2F' F' diminished
n= (≥ 1) I
13 Focal length of lens is f. Lens formula:
sin 𝜃m
1 1 1
no unit = +
f u v
Beyond 2F'
4 Refractive index n is also related to the speed of light c in a u: positive
vacuum and v in the medium: v: positive for real, negative for virtual
c between real,
n=v (≥ 1) O F 2F F and inverted, f: positive for convex, negative for concave
2F' F' 2F diminished
5 General form of Snell’s law: n1 sin θ1 = n2 sin θ2 I
14 Graphs (for convex lens)
n1 < n2 n1 > n2
1
v v
θ1 At 2F'
θ1
n1 n1
n2 u = v = 2f
n2
θ2 O real, 1 slope = –1
θ2 F 2F at 2F inverted, f
2f
2F' F' I same size 1
u
u 2f
6 Conditions for total internal reflection:
(1) Light travels from larger n to smaller n.
Between F' and 2F'
(2) i > C

real,
7 Critical angle C of a medium and its refractive index n: O F 2F beyond
inverted,
1
2F
C = sin–1 ( ) 2F' F' I magnified
n

8 Construction rules for lenses

Convex lenses Concave lenses At F'

O F 2F at
no image
F' F F' F infinity
2F' F'

Within F'
F' F F' F
same virtual,
I F 2F
O side as erect,
2F' F' object magnified

F' F F' F

1
IIIb Wave Motion 12 Refraction of waves 18 Nodal and antinodal lines 24 Degree of diffraction of light:
red light > green light > blue light
A N A N A N A N A
1 Amplitude A and wavelength λ:  wavelength of red > green > blue
travel direction
b f 25 Interference of light:
A λ Interference pattern in Young’s double slit experiment
a c e g consists of a series of evenly spaced dark and bright
From deep to shallow From shallow to deep
d fringes .
2 Particles c, g are moving upwards;
i i 5 4 3 2 1 ∆=0 1 2 3 4 5
particles a, e are moving downwards;
particles b, d, f are momentarily at rest. r
r
19 Nodal/antinodal lines spacing  if  or
3 Frequency f and period T: separation between sources 
1
f=
T
9 7 5 3   3 5 7 9
Unit: Hz s f unchanged Sources are 1 apart: Sources are 3 apart:
2 2 2 2 2 2 2 2 2 2

v    
0 0
4 Particle vibrates once  wave moves forwards by 1λ 3  2 2  3
Bright fringes = constructive interference
     2 2
2 2 2 2
5 Wave speed v, frequency f and wavelength λ 5 5
(path difference = 0, 1, 2, 3, …)
bends towards normal bends away from normal
2 2
v = f (i > r) (i < r) Dark fringes = destructive interference
Unit: m s –1   1 1 1
Hz m 3 3 (path difference = , 1 , 2 , …)
13 Speeds in media 1 and 2 are v1 and v2. 2 2 2

6 Wave speed depends on medium Wavelengths in media 1 and 2 are 1 and 2.
Refractive index from medium 1 to medium 2 (n12): 20 Formation of stationary waves (standing waves): 26 Monochromatic light = light of single wavelength
(for springs: tension T and mass per unit length μ )
v1 1 A stationary wave forms when two waves of the same Example: laser
n12 = =
v2 2
7 In phase and antiphase frequency and amplitude travelling at
27 Fringe separation y on a screen at distance D from slit (slit
14 Diffraction of waves
Particles a and e are in phase. the same speed in opposite directions . separation a) and wavelength of light :
Particles a and c/g are in antiphase.
21 Nodes and antinodes of stationary wave
8 Water waves projected on screen
a Draw two adjacent wavefronts A A A A
b Draw direction of travel N N N N N
c Indicate wavelength

22 Compare travelling wave and stationary wave:

15 For diffraction at a gap, degree of diffraction  if Travelling wave Stationary wave

λ λ
gap width  or 
Move or not Moves forwards Does not move

straight wave circular wave 16 Locating positions of constructive and destructive Different for D
Amplitude Same for all particles y =
interference different particles a

Phase difference Path difference Unit: m m m m

9 Speed v of water waves in ripple tank depends on Same loop: in
Constructive Two waves arrive  = 0, 1, 2, Out of phase for phase Assumption: D >> a
Phase relation
water depth . interference in phase 3 , … neighbouring particles Adjacent loop: out
(water vibrates (e.g. crest meets of phase
crest , Fringe separation  if
10 For constant water depth, v is constant , violently )
trough meets Different particles
f   λ  Time to reach
reach smax at different
All reach their smax (i) slit separation  ;
trough ) smax at same time
11 Reflection of waves ( r = i ) Destructive Two waves arrive 1
 = , 1 ,
1 times (ii) wavelength of light  ;
interference in antiphase 2 2
1
(e.g. crest meets 2 , … Energy From one place to (iii) distance from screen 
barrier (water stays 2 Stored in wave
trough , transmission another
u
calm ) trough meets 28 Angle of fringe  of n-th order bright fringe:
v crest )
50° 50° 23 Light is a wave because it shows diffraction a sin θ = n
17 Coherent sources: same frequency and and interference . Unit: m m
barrier u=v
constant phase difference
 steady and observable interference pattern
1
29 Plane transmission grating = many many slits (e.g. 600 slits
per mm)
Interference pattern in plane transmission grating:

2nd 1st 0th order 1st 2nd not evenly

spaced

The fringes formed by a grating are brighter and more

widely spaced .

30 Angle of fringe  of n-th order bright fringe (grating spacing d,

wavelength of light ):
d sin θ = n
Unit: m m

Fringe separation  if
(i) grating spacing  ; (fewer lines per mm)
(ii) wavelength of light  ;
(iii) distance from screen 

31 Highest-order of bright fringe:

d
nmax ≤


32 Electromagnetic spectrum
frequency

radio infra-red ultra-violet gamma

radiation radiation
X-rays
waves rays

microwaves visible light

wavelength

33 Visible light: 400 nm (violent) to 700 nm (red)

34 Ultra-violet, X-rays, gamma rays: higher frequency, higher

energy

35 Longitudinal wave:
(i) vibration of particles along direction of travel of wave
(ii) consists of compression and rarefaction

36 Wavelength :

0 1 2 3 4 5 67 8 9 10 11 12 13 14

d


37 Audible frequency range: from 20 Hz to 20 kHz

38 Sound intensity level: measures loudness/energy of

sound
Unit: decibel ( dB )
Softest sound heard by human ear = 0 dB

2
IV Electricity and Magnetism 6 Positive charge q experiences electric force F at a
Voltmeter
Voltage (conservation of energy )
position. Electric field strength E at that position:
1 Model of an atom: F
E= Resistor V1
q
Unit: N C–1 C N ε ε V1 V2
proton Variable resistor / rheostat
neutron V2
7 Electric field strength E at distance r from point charge Q:
Earthed point / earth wire
1 Q Q ሬሬԦ
E= • r E
electron 4π𝜀0 r2 Thermistor (temperature- ε = V1 + V2 ε = V1 = V2
Unit: N C–1 m C dependent resistor)

2 Charge of particles: 8 Electric field strength E between parallel charged plates A.c. source Definition of resistance: V = IR
–19
Electron: –1.60 × 10 C = –e (p.d. V) separated by distance d: Resistance in series (R1, R2,… in series): R = R1 + R2 + …
Fuse
–19 V +
Proton: +1.60 × 10 C = +e V 1 1 1
E=
d m Resistance in parallel (R1, R2,… in parallel): = + +…
R R1 R2
Neutron: 0 15 Variation of I with applied p.d. V for different materials:
Unit: N C–1 or V m–1 d V
ሬሬԦ
E
Metal wire 20 E.m.f.  and terminal voltage V of a power source with
3 Electric field patterns:
– I internal resistance r:
Isolated charge (+): Isolated charge (–):
9 Direction of electric force on a charge in electric field: ε r
+ charge: along E-field +q F
straight line
V
 charge: opposite to E-field ε = V + Ir
E –q V
F V = IR
pass through origin

10 When there is charge Q flowing through in time t, electric Filament lamp

current I: I 21 Measure resistance of resistor:
(1) radially outwards / inwards
Q If resistor has high resistance: If resistor has low resistance:
(2) farther away  field lines more widely spread  weaker field I=
t

Electric field pattern between two oppositely charged parallel Unit: A s C (A = C s–1)

plates: +Q 11 Potential difference (p.d.) between two points in external V

circuit where charge Q loses electric potential energy E:
(1) from +ve to –ve
E
p.d. = Q
(2) straight, parallel, evenly
spaced Unit: V C J (V = J C–1) 16 Ohm’s law (a special case of resistance behaviour)
 uniform field –Q
VI
12 Electromotive force (e.m.f.)  of battery that increases the
4 Electric force F between two charges Q1 and Q2 separated by Assumption: T, etc. are unchanged
electric potential energy of charge Q by E:
distance r :
E
1 Q1 Q2 ε=Q 17 Resistance R of metal wire of length l, cross section area A
F = 4π𝜀 • r2
(Coulomb’s law) and resistivity : 22 Electrical power P of component when energy E is converted
0
Unit: V C J in time t:
Unit: N m C l
R=ρ E
A P=
0 (permittivity of free space) = 8.85  10–12 C2 N–1 m–2 13 Resistance R (definition) of a component with voltage V and t
current I: Unit: Ω Ωm m 2
m
V Unit: W s J
5 Adding electric forces: V
R= I 18 Effect of temperature on resistance
I
as T 
Formulae relating P, V, I and R:
two opposite charges (+Q Unit: Ω A V (Ω = V A–1) R Metals: R 
two same charges (+Q):
and –Q): Semiconductors (e.g. silicon ): R  as T 
2
V R I
P = VI = I2R =
R
14 Circuit symbols
19 Circuit analysis: V
F Circuit component Circuit symbol
Current (conservation of charge ) 23 For the same type of light bulb,
θ θ
Cell higher power  brighter
+q +q I1 I2 I3
F (1) I1 = I2 = I3
r r Battery 24 Electrical energy E expressed in kilowatt-hour:
r r
θ θ I1 I2
Wires at a joint (2) I1 = I2 + I3
I3
E = Pt
+Q –Q +Q +Q
Switch Unit: kW h kW h
1 Qq 1 Qq
F = 2(4πε ∙ r2
) cos θ F = 2(4πε ∙ r2
) sin θ
0 0
Light bulb/lamp
(to the right) (upwards)
Ammeter

1
25 Mains electricity in HK: 30 A straight conductor carries current I. It has a part of length l 34 A coil of area A lies in magnetic field B. Angle between 39 Transformer:
lying within magnetic field B and making angle  with field. normal of coil and field is .
voltage: 220 V , frequency: 50 Hz Ip Is
Magnitude of magnetic force F on the conductor: Magnetic flux  through the
coil:
26 Connection to appliance F = BIl sin θ
 = BA cos θ
 Live wire (insulation colour: brown ): Unit: N T A m Vp Vs
(T = N A–1 m–1) Unit: Wb T m2
varies between +311 V and –311 V
periodically.
 Neutral wire ( blue ): Magnetic flux linkage (N turns) = N Np Ns
always stays at 0 V.
 Primary voltage Vp, secondary voltage Vs
 Earth wire ( green and yellow ): Magnetic flux density  magnetic field B =
A Primary current Ip, secondary current Is
Direction: Fleming’s left hand rule
connects appliance body to earth . Unit: Wb m–2 m2 Wb Number of turns of primary coil Np
Number of turns of secondary coil Ns
F
27 Fuse and switch should be connected in live wire. 35 EM induction: For a magnetic flux change  in time t,
induced e.m.f. : Vp
Voltage ratio and turns ratio: V = N
Np
s s
28 Representation of direction: ε=
∆
(Faraday’s law)
∆t
Current and voltage (efficiency ):
B Unit: V s Wb
VpIp × η = VsIs
36 Direction: Fleming’s right hand rule 40 Power transmission:
F I
I
out of the page into the page
B Vin r Vout
31 An N-turn coil with area A carries current I. It lies within
29 Magnetic field due to currents magnetic field B, with its normal making angle  with field. I
a. Long straight wire: Torque  on the coil:
Magnitude of field B at
Total resistance of cables = r
distance r from wire
carrying current I: I Current in cables = I

Power loss P in cables = I2r

𝜇0 I
B = 2𝜋r Or Lenz’s law: Voltage drop V in cables = Ir
Unit: T m A Induced e.m.f. (current) opposes the change causing it.

 = BIAN sin θ 37 Effect value of a.c. = root-mean-square value of a.c.

̅
Average power P = VrmsIrms
(1) circles Unit: Nm T A m2
(2) farther away  more widely
spaced  weaker field 38 For sinusoidal a.c. with peak current I0 and peak voltage V0:
32 A charge Q moves at speed v in magnetic field B. Its moving
0 (permeability of free space) = 4  10–7 T m A–1 direction makes angle  with the field. Magnitude of magnetic
I
force F on the charge:
b. Long solenoid: I0
Irms
Magnitude of field B inside F = BQv sin θ = 0.707I0
solenoid of N turns, length l Unit: N C
and carrying current I: T m s–1

𝜇0 NI
B=
l
Unit: I0
T m A Irms =
√2

straight line, evenly spaced V0

Vrms =
 uniform field (inside) Work done by B-field on moving charge = 0 √2

33 A particle of charge Q and mass m moves in direction

c. Circular coil (pattern only): perpendicular to magnetic field B with speed v. Its path would
be a circle of radius r:
straight at mid-point
kg
mv2
BQv =
r
Unit: T m s–1 m s–1
C m

2πr 2π
T= v
= 𝜔

2
V Radioactivity and Nuclear Energy 7 Half-life 𝑡1 :
2

ln 2
1 Compare different types of nuclear radiation: ,  and : t1 =
2
k
 radiation  radiation  radiation
8 Mass m converted to energy E:
Nature He nuclei e– EM waves
Electric ∆E = ∆mc2 (mass-energy relationship)
+2e –e 0
charge
Unit: J kg m s–1
Speed various; < c various; < c =c
Ionizing 9 Atomic mass unit u:
α > β > γ
1 u = 12 × C-12 atom = 1.661  10–27
power 1
kg
Stopped by Stopped by
Penetrating Halved by
power
1–2 sheets of 5-mm
25-mm lead 1u= 1.493  10–10 J = 931 MeV
paper aluminium
Range in
several cm several m > 100 m 10 Electron-volt:
air
1 eV = 1.60  10–19 J

In E-field

In B-field

Cloud
chamber
tracks
thick and thin and scattered,
straight twisted hardly seen

2 Equation of alpha decay:

mass no.
A
ZX → A–4 4
Z–2Y + 2He
atomic no.
3 Equation of beta decay:
A
ZX → Z+1AY + 0
–1e

4 Equation of gamma decay:

with higher energy

A
ZX* → AZX +γ

5 Activity A of a sample and number of undecayed nuclei N:

A = kN
Unit: Bq (s–1)

k: decay constant; unit: s–1/min–1/day–1

6 Number of undecayed nuclei N after time t:

N = N0e–kt
Unit: s–1 s