Written by students who passed Immediately available after payment Read online or as PDF Wrong document? Swap it for free 4.6 TrustPilot
logo-home
Document preview thumbnail
Preview 1 out of 4 pages
Other

Sheet of All A-level Physics Equations

Document preview thumbnail
Preview 1 out of 4 pages

This is a document of all of the equations, constants, and other important number related information points. The equations are organised neatly in their corresponding topic headings and colour coded for the sections they are in. Each equation has its units next to it and an arrow beside the equation is used to denote which equations and constants you are given in the AQA data sheet. Even if you are not taking the AQA course, a lot of the equations will likely be in both of the courses anyway, just ignore the arrows.

Content preview

P1 - Matter and Radiation P2 - Quarks and Leptons
Particle Charge (C) Rel. Charge Mass (kg) Rel. Mass Particle/ Rel. Rest Energy Interaction
Antiparticle Charge (MeV)
-19 -27
Proton +1.6 x 10 1 1.673 x 10 1 Photon γ 0 0 None
Neutron 0 0 1.675 x 10 -27
1 Neutrino ν 0 0 Weak
Antineutrino ν̅
Electron -1.6 x 10-19 -1 9.11 x 10-31 0.0005 Electron e- -1 0.510999 Weak, electromagnetic
charge Positron e+ +1
Specific Charge = mass (C kg-1)
Muon µ- -1 105.659 Weak, electromagnetic
Electron Specific Charge = 1.76 x 1011 C kg-1
Antimuon µ+ +1
Proton Specific Charge = 9.58 x 107 C kg-1 Pions π±, π0 ±1, 0 139.576 (±), Strong, electromagnetic, weak
1 eV = 1.60 x 10-19 (J) π+ for a π- ±1, 0 134.972 (0)
Atomic Mass Unit, u = 1.661 x 10-27 kg = 931.5 MeV π0 for a π0
Planck Constant, h = 6.63 x 10-34 (J s) Kaons K±, K ±1, 0 493.821 (±), Strong, electromagnetic, weak
8 -1 K+ for a K- ±1, 0 497.762 (0)
Speed of Light, c = 3.00 x 10 (m s )
K0 for a K0
Radio Microwave IR Visible UV X-rays Gamma Proton p +1 938.257 Strong, weak, electromagnetic
> 0.1 m 0.1 m - 1 mm - 700 nm - 400 nm - 10 nm - < 1 nm Antiproton p̅ -1
700 nm 400 nm 1 nm 0.001 nm Neutron n 0 939.551 Strong, weak
1 mm
Antineutron n̅
Wavelength, λ = c (m)
f
hc Property Quarks Antiquarks
Photon Energy, E = hf = (J)
λ
Power of Photon Beam of n Photons per Second, P = nhf (W) Flavour Up u Down d Strange s Up u̅ Down d̅ Strange s̅
Charge Q +2/3 -1/3 -1/3 -2/3 +1/3 +1/3
Strangeness S 0 0 -1 0 0 +1
P3 - Quantum Phenomena
Baryon No. B +1/3 +1/3 +1/3 -1/3 -1/3 -1/3
Energy of a Photoelectron, E = hf = EKmax + φ (J)
φ Lepton Lepton
Threshold Frequency, fmin = (Hz)
h Symbols Number
I -1
Photoelectrons from Cathode to Anode per Second, n = (s )
q Particles e-, νe : μ-, νμ +1
Energy Levels, E = hf = E1 - E2 (eV) + +
Antiparticles e , ν̅e̅ : μ , ν̅μ̅ -1
Hydrogen Energy Level = - 13.62 eV (eV)
n h
De Broglie Wavelength, λ = h = (m)
p mv

P5 - Optics
P4 - Waves Refractive Index of Substance s, n = c
c
s
sin i λ
Refractive Index of Substance s, n = =
Frequency, f = 1 sin r λs
T
1 Snell’s Law of Refraction, n1sinθ1 = n2sinθ2
Time Period, T = (s) n2
f
Critical Angle, sinθc = n for n1 > n2
Wave Speed, c = fλ (m s-1) λD 1
2πd c Fringe Spacing, w = s (m)
Phase Difference = ()
λ λD
Slit Separation, s = (m)
Distance Between Adjacent Nodes = λ (m) w
2
T Double Slit Reinforcement Path Difference = mλ (m)
First Harmonic Frequency, f = 1 (Hz)
2l μ 1
Double Slit Cancellation Path Difference = (m + 2 )λ (m)
2Dλ
Width of Central Fringe, W = a (m)
Diffraction Grating of nth Order, nλ = dsinθ
P6 - Forces in Equilibrium 1
Slits Per Metre, N = (m-1)
d
Magnitude of Resultant of Two Perpendicular Forces, F = (F12 + F22) (N)
F
Angle Between Resultant and F1, tanθ = F2
1
Force Parallel to the Line = Fcosθ (N) P7 - On The Move
Force Perpendicular to the Line = Fsinθ (N) Δs
Speed, v = (m s-1)
Support Force From Object Resting on Horizontal Plane, S = W (N) Δt
2πr
Speed in a Circle, v = (m s-1)
Reaction/Support Force on Sloped Plane, R/S = Wcosθ (N) T
Δv v - u
Acceleration, a = = (m s-2)
For a System in Equilibrium, F1 + F2 + … + Fn = 0 Δt Δt
-1
Final Velocity, v = u + at (m s ) v2 = u2 + 2as (m2 s-2)
Moment of a Force, M = Fd (N s) (v + u)t 2
Displacement, s = = ut + at (m)
For Moments in Equilibrium, F1d1 = F2d2, M1 = M2 (N s) 2 2 2
Displacement, s = vt - at (m)
For Moments Involving Centre of Mass of Beam, W0d0 = W1d1 (N s) 2
Horizontal Component of Projectile Displacement, x = utcosθ (m)
Support Force at Pivot, S = F0 + F1 + F2 (N) gt2
Vertical Component of Projectile Displacement, y = utsinθ - (m)
2
Horizontal Component of Projectile Velocity, vx = ucosθ (m s-1)
Vertical Component of Projectile Velocity, vy = usinθ - gt (m s-1)

Connected book
 image
Jim Breithaupt AQA Physics: A Level
Publisher: 2016 ISBN: 9780198378495 Edition: Unknown

Document information

Study Level
Subject
Uploaded on
March 5, 2023
File latest updated on
March 18, 2023
Number of pages
4
Written in
2022/2023
Type
Other
Person
Unknown
$10.44

Wrong document? Swap it for free Within 14 days of purchase and before downloading, you can choose a different document. You can simply spend the amount again.
Written by students who passed
Immediately available after payment
Read online or as PDF

Sold
0
Followers
0
Items
1
Last sold
-



Why students choose Stuvia

Created by fellow students, verified by reviews

Quality you can trust: written by students who passed their exams and reviewed by others who've used these revision notes.

Didn't get what you expected? Choose another document

No problem! You can straightaway pick a different document that better suits what you're after.

Pay as you like, start learning straight away

No subscription, no commitments. Pay the way you're used to via credit card and download your PDF document instantly.

Student with book image

“Bought, downloaded, and smashed it. It really can be that simple.”

Alisha Student

Working on your references?

Create accurate citations in APA, MLA and Harvard with our free citation generator.

Working on your references?

Frequently asked questions