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VIJAYA COMPOSITE PU COLLEGE South End Circle, Jayanagar III Block, Bangalore-560011
Website:www.vijayainstitutions.org
REFRESHER COURSE MATERIAL FOR II PUC
ACADEMIC YEAR 2019-20
SUBJECT: PHYSICS
UNITS AND DIMENSIONS
PHYSICAL QUANTITY DIMENSIONS SI UNITS
LENGTH [L] m
AREA [L2] m2
VOLUME [L3] m3
MASS [M] kg
DENSITY [ML-3] Kgm-3
TIME [T] S
VELOCITY [LT-] ms-1
ACCELERATION [LT-2] ms-2
FORCE [MLT-2] N
WORK [ML2T-2] J
ENERGY(TYPES) [ML2T-2] J
POWER [ML2T-3] W
CURRENT [A] A
MOMEMUNTUM [MLT-1] Kgms-1
IMPULSE [MLT-1] Ns
PRESSURE(STRESS) [ML-1T-2] Nm-2 or pascal
STRAIN NO DIMENSION No unit
TORQUE [ML2T-2] Nm
SPECIFIC HEAT CAPACITY
[M0L2T-2K-1] Jkg-1k-1
MOTION IN ONE DIMENSTION
MOTION : Change in position of an object wrt time and surroundings.
TRANSLATORY MOTION : The linear distance covered by a particle
PATH LENGTH (DISTANCE) : The distance between two given points
DISPLACEMENT : The shortest distance between two points
DIFFERENCE BETWEEN DISTANCE AND DISPLACEMENT
DISTANCE DISPLACEMENT
Actual length traressed by an object Shortest distance between two points
Scalar quantity Vector quantity
It can never be negative or zero Can be positive negative or zero
Speed : Rate of change of distance covered by the particle
Speed = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑡𝑖𝑚𝑒 ( SI unit → ms-1)
TYPES OF SPEED
(a) Uniform speed : If an object describes equal distance in equal internals
of time , however the small the internal may be
(b) Non- uniform speed : Equal distance in unequal interval of time or vice
Versa
(c) Average speed : The ratio of the total distance covered by the body to
the total time taken
Avg . speed Vavg = 𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒
VELOCITY : rate of change of displacement
V= 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑡𝑖𝑚𝑒 ms-1 vector quantity
TYPES
Uniform velocity : If a body covers equal displacements in equal
internals of time , however small the time internal
Non – uniform velocity : If a body covers un equal displacement in
equal interval of time or vice versa
however small the time internal may be
Average velocity : The ratio of the displacement to the time
interval of which the motion taken place
Acceleration : Rate of change of velocity
a = 𝑉−𝑢
𝑡 =
∆𝑣
∆𝑡 ms-2 vector
EQUATIONS OF MOTION FOR A UNIFORMLY ACCELERATED MOTION
1) v =u+ at 2) S = ut + 1
2 at2 3) v2 = u2 +2as
MOTION IN A PLANE
PHYSICAL QUANTITY : Any quantity which can be measured is called physical
quantity
SCALAR : A physical quantity which has only magnitude and direction
Eg : L ,M, V density etc
VECTOR : A physical quantity which have both magnitude as well as direction
Eg : velocity , acceleration, impulse, force etc
PARALLELOGRAM LAW OF VECTOR ADDITION
RESOLUTION OF VECTORS
The process of splitting of a vector into two or more components is called
resolution of vectors
RECTANGULAR COMPONENTS OF A VECTOR
Resolving a single vector into two rectangular components
UNIFORM CIRCULAR MOTION
The motion of an object along a circular path with constant angular speed
CENTRIPETAL ACCELERATION : The acceleration of the object undergoing an
uniform circular motion and directed along the radius of the circle and towards
its centre and towards its centre its centre is called centripetal acceleration .
Relation between LINEAR VELOCITY and ANGULAR VELOCITY
V = r w
Relation between LINEAR ACCELERATION and ANGULAR ACCELERATION
A = r 𝛼
Expression CENTRIPETAL FORCE
F = 𝑚𝑣2
𝑟 and a =
𝑣2
𝑟 (cenp…. Acc)
1. Two forces of magnitude 6N and 8N are inclined at angle of 30° and 60°
With the horizontal find the resultant force.
F1 = 6N , 𝜃1 = 30°
F2 = 6N , 𝜃1 = 60°
F1 → resolved into 2 components
F1x = F1 cos 𝜃1
F1y = F1 sin 𝜃1
F1x = 6 x cos30 = 6 x √3
2 =3√3 N along x axis
F1y = 6 x sin30 = 6 x 1
2 = 3N along y axis
F2x = F2 cos 𝜃2 = 8 x cos60 = 8 x 1
2 = 4N along x axis
F2y = F2 cos 𝜃2 = 8 x sin60 = 8 x√3
2 =4√3 N along y axis
Net Fx = F1x + F2x
Fx = (3√3 +4)N
Fy = F1y + F2y
Fy = (3+4√3) N
F= √𝑓𝑥2 + 𝑓𝑦2
= √(3 √3 + 4) 2 + (3 + 4√3) 2
= √27 + 16 + 24√3 + 9 + 48 + 24√3
= √100 + 48√3 = 13.531N
2. A force of 10N is acting at an angle of 120° with the x- axis. Determine
the X & Y components of the force .
Fx = F cos𝜃
= 10 x cos60
= 10 x 1
2 = 5N
5N along – x axis (xI – axis)
Fy = F sin𝜃
= 10 x sin60
= 10 x √3
2 = 5 √3 N
= 5 √3 N along y – axis
ENERGY : capacity to do work is known as energy
TYPES OF MECHANICAL ENERGY
POTENTIAL ENERGY : Energy possessed by a body due to its position
configuration is PE
Gravitation PE = mgh
KINETIC ENERGY : Energy possessed by a body by virtue of motion is called
kinetic energy
KE = 1
2 mv2
NEWTONS II LAW
It states that, rate of change of momentum of a body is directly proportional
To the applied and acts in the direction of applied force .
�� = m��
NEWTONS III LAW
To every action ,there is equal and opposite reaction
Eg : walking , swimming
𝐹𝐴𝐵 = - 𝐹𝐵𝐴
SCALAR PRODUCT OF TWO VECTORS
𝐴 . �� = AB cos𝜃
VECTOR PRODUCT OF TWO VECTORS
𝐶 = 𝐴 X ��
�� X �� = |��| |��| sin𝜃 ��
�� → direction normal to the plane containing �� & ��
WORK : Product of force and displacement
W = �� . 𝑆 = FS cos𝜃 S.I UNIT is Joule
POWER : Rate of doing work
P = 𝑊
𝑇 unit watt (W)
1 watt = 1 JS-1 , 1kw =1000 w
1MW = 106 W
1 Kwh = 3.6 x 106 J , 1HP = 746W
TORGUE : The turning effect of force produced in a body about a fixed axis
�� = 𝑟 x �� = r F sin𝜃 . ��
ANGULAR MOMENTUM: Moment of linear momentum
�� = 𝑟 x �� = r p sin𝜃 . ��
OSCILLATIONS
PERIODIC MOTION : Any motion possessed by the body if repeats it self after
regular intervals of time is called periodic motion
Eg : Rotation of earth about its own axis
Revolution of earth around the sun
OSCILLATORY MOTION : the motion in which a body moves to and fro or back
and fouth about the fixed position called mean position in a definite interval of
time
Eg : pendulum of the wall clock
PERIOD : Time taken to complete one to and flo motion .
S.I unit is second (s)
FREQUENCY : number of oscillations made by the particle in one second
T = 𝐼
𝑓 S I unit of frequency is Hertz (Hz)
AMPLITUDE : Maximum displacement of a oscillating particle on either side of
its mean position
DISPLACEMENT EQUATION FOR SHM
y = A sin (wt + ∅) x = A cos (wt+∅)
GRAPHICAL REPRESENTATION
PRINCIPLE OF SUPERPOSITION
“ The resultant displacement any particle of the medium is equal to the
algebraic sum of individual displacement “
�� = 𝑦1 + 𝑦2
PHASE : The state of vibration of a particle at that instant
ELECTRIC CHARGES AND FIELDS
ELECTRIC CHARGE : it is an intrinsic properly which is responsible for electric
force of attraction repulsion between two objects
S.I unit of electric charge is coulomb (C)
TYPES OF CHARGE : 1) positive charge 2) negative charge
The body which loses electrons is said to be positively charged
The body which gains the electrons is said to be negatively charged
GOLD LEAF ELECTROSCOPE :
Gold leaf electroscope is used to detect the pressure of electric
charge on a body .
BASIC PROPERTIES OF ELECTRIC CHARGE :
1) Like charges repel each other
2) Unlike charges attracts each other
3) Additivity of charges
4) Charge is conserved
5) Quantisation of charge
ELECTRIFICATION
The process of adding and removing an electron from a surface of a body
TYPES OF ELECTRIFICATION
1) Charging by conduction 2) Charging by friction 3) Charging by induction
COULOMB’S LAW :
The magnitude of force between two point charges is directly proportional
to the product of the magnitude of the charges and inversely proportional to
square of the distance between them .
q1,q2 – point charges , r ---- distance between two charges
F 𝛼 q1 q2→ 1 F 𝛼 1
𝑟2 → 2
From equation 1 and 2
F 𝛼 𝑞1𝑞2
𝑟2 , F = 𝑞1𝑞2
𝑟2 where K = 1
4𝜋∈𝑜 = 9 x 109 Nm2c-2
F = 1
4𝜋∈𝑜 𝑞1𝑞2
𝑟2 K is the electrostatic force constant
∈𝑜 permittivity of free space
VECTOR FORM OF COULOMB’S LAW
�� = 1
4𝜋∈𝑜 𝑞1𝑞2
𝑟2 �� �� → is the unit vector
DEFINITION OF 1 COULOMB :
One coulomb is the charge that when placed at a distance of 1m from another
Charge of the same magnitude in vaccum experiences an electrical force of
repulsion of magnitude 9 x 109 N .
If q1 = q2= 1C r = 1m 1
4𝜋∈𝑜 = 9 x109
F = 1
4𝜋∈𝑜
𝑞1𝑞2
𝑟2 = 9 x 109 1𝑥1
12 = 9 x 109 N
ELECTRIC FIELD :
The space around a point charge where a unit positive charge experiences a
force
ELECTRIC FIELD INTENSITY
The force experienced by a unit positive charge placed at that point
�� = ��
𝑞 S.I unit N/C or V/m It is a vector quality
SUPERPOSITION PRINCIPLE
Total force acting on a given charge due to number of charges is the vector
sum of the force due to individual charges acting on that charge
ELECTRIC FIELD LINES
Electric field line is the path along which a free unit isolated
positive charge tend to moves
The direction of electric field due to a point positive charge is
radially outward
The direction of electric field due to a negative point charge is
radically inward .
PROPERTIES OF ELECTRIC FIELD LINES
Electric field lines originate form positive charge and terminate on the
negative charge
Electric field lines do not intersect
They do not form any closed loops
Electric field lines are always perpendicular to the surface of the
conductor
Electric field lines do not pass through the conductor
ELECTRIC FLUX
The number of electric field lines crossing unit area of cross section normally
∅ = ∑ �� . ∆ 𝑆 = ∑ E ∆ S cos𝜃 / ∆ ∅ = E . ∆S = E∆S cos𝜃
It is a scalar quality ∆∅ → electric flux
S.I unit of electric flux is NC-1 m2 ∆𝑆 → an area of element
ELECTRIC DIPOLE
A pair of equal and opposite point charges separated by a small distance
Ex : water ammonia Hcl and CO
ELECTRIC DIPOLE MOVEMENT
The product of magnitude of either charge and the distance between them
�� = q x 2a �� 2a → dipole length
P = q x 2a �� → unit vector along the dipole axis
The direction of �� is form –q to +q along the dipole axis
S.I unit is C-m (coulomb meter)
It is vector quantity
THE ELECTRIC FIELD AT A POINT ON THE AXIS OF A DIPOLE
E+ = 1
4𝜋∈𝑜
𝑞
(𝑟−𝑎) 2 �� and E _ =
1
4𝜋∈𝑜
−𝑞
(𝑟+𝑎) 2 ��
Where �� is a unit vector along the axis of the dipole
E = E+ + E-
= 1
4𝜋∈𝑜
𝑞
(𝑟−𝑎) 2 �� -
1
4𝜋∈𝑜
−𝑞
(𝑟+𝑎) 2 ��
= 𝑞
4𝜋∈𝑜 [
1
(𝑟−𝑎) 2−
1
(𝑟+𝑎) 2] ��
E = 𝑞
4𝜋∈𝑜
4𝑟𝑎
(𝑟2−𝑎2) 2 ��
E = 1
4𝜋∈𝑜
𝑞 𝑋 2𝑎 .2𝑟
𝑟4 �� for r ≫ a
�� = 1
4𝜋∈𝑜
2��
𝑟3 �� = 𝑞 𝑋 2𝑎 ��
P = q X 2a dipole movement
THE ELECTRIC FIELD AT A POINT ON THE EQUILATERAL LINE OF A DIOPLE
E + = 1
4𝜋∈𝑜
𝑞
(𝑎2+𝑟2)
2𝑞
4𝜋∈𝑜 +
1
(𝑎2+𝑟2)
E - = 1
4𝜋∈𝑜
𝑞
(𝑎2+𝑟2)
Cos𝜃 = 𝑂𝐵
𝑃𝐵 =
𝑎
√𝑎2+𝑟2 =
𝑎
(𝑎2+𝑟2)1
2
The total electric field is opposite to ��
E = - (E+q + E-q) cos𝜃 ��
= - [1
4𝜋∈𝑜
𝑞
(𝑎2+𝑟2)+
1
4𝜋∈𝑜
𝑞
(𝑎2+𝑟2) ]
𝑎
(𝑎2+𝑟2)1
2
��
= [2𝑞
4𝜋∈𝑜 +
1
(𝑎2+𝑟2)]
𝑎
(𝑎2+𝑟2)1
2
��
= - 1
4𝜋∈𝑜
𝑞 𝑋 2𝑎 ��
(𝑎2+𝑟2)3
2
�� = - 1
4𝜋∈𝑜
��
𝑟3 for r ≫≫ 𝑎 �� = q X 2a ��
p = q x 2a P → Dipole moment
�� = 1
4𝜋∈𝑜
2��
𝑟3
THE TORQUE ACTING ON AN ELECTRIC DIPOLE PLACED IN AN UNIFORM
ELECTRIC FIELD
𝜏 = force x perpendicular distance
𝜏 = qE X BC = qE (2asin𝜃)
𝜏 = q x 2a E sin𝜃
𝜏 = PE sin𝜃
Sin 𝜃 = 𝑜𝑝𝑝
ℎ𝑦𝑝 =
𝐵𝐶
2𝑎
BC = 2a sin𝜃
Continuous charge distribution
Linear charge density (⋋) : The amount of charge per unit length
⋋ = ∆𝜃
∆𝑙 , cm-1
Surface charge density (𝜎) : The amount of charge per unit area
𝜎 = ∆𝜃
∆𝑆 cm-2
Volume charge density (ƍ) the amount of charge per unit volume
ƍ = ∆𝜃
∆𝑆 cm-3
Polar molecules : The centre of positive and negative charges do not coincide
hence have a permanent electric dipole movement
Ex H2O (water)
Non – polar molecules : the centre of positive and negative charges lie at the
same plane and have zero dipole movement
Ex CO2 ,CH4
State Gauss’s law in electrostatics
The total electric flux through a closed surface is equal to 1
∈𝑜 times the total
charge enclosed by that surface
∅ = 𝑞
∈𝑜
The electric field due to straight infinity long uniformly charged wire using
gauss law
L – length r – radius
Flux through two ends of the cylinder is zero because field is radial flux through
curved cylindrical part
∅ = E x 2𝜋 rl cos 𝜃 = E x 2𝜋 rl ① 𝜃 = 0°
⋋= linear charge density ⋋= 𝑞
𝑙
q = ⋋l
from gauss law ∅ = 𝑞
∈𝑜 =
⋋l
∈𝑜 ②
from eqn ① and ②
E x 2𝜋 rl =⋋l
∈𝑜
E = ⋋
2𝜋𝑟∈𝑜
�� = ⋋
2𝜋𝑟∈𝑜 �� where �� is a unit vector
Electric field due to uniformly charged infinite plane sheet using gauss law
∅ = E x area of the end face of the cylinder
∅ = E .2A ①
Gauss theorem
∅ = 𝑞
∈𝑜 𝜎 =
𝑞
𝑎
∅ = 𝜎𝐴
∈𝑜 ② q = 𝜎 A
From equation ① and ②
E X 2A = 𝜎𝐴
∈𝑜
E = 𝜎
2∈𝑜 or �� =
𝜎
2∈𝑜 ��
Electric field due to uniformly charged thin spherical shell when the point is
outside the conductor
Electric flux d∅ = (E cos𝜃) (ds)
d∅ = (E cos𝜃) (ds)
d∅ = E ds
for the entire surface
∅ = E ∫ 𝑑𝑠
= Es
∅ = E. 4𝜋𝑟2 → ① where S is the surface area = 4𝜋𝑟2
According to gauss law
∅ = 1
∈𝑜 q → ②
From ① and ②
E. 4𝜋𝑟2 = 𝑞
∈𝑜
E = 1
4𝜋∈𝑜
𝑞
𝑟2
1) Two point charges qA = 3𝜇𝑐 and qA = - 3𝜇𝑐 are located 20cm apart in
vaccum
a) What is the electric field at the mid point 0 of the line AB joining the
two charge ?
b) If a negative test charge of magnitude 1.5 X 10-9 C is placed at this
point , what is the force experienced by the test charge ?
EA = 1
4𝜋∈𝑜
𝑞𝐴
𝑂𝐴2 = 9 𝑋 109 𝑋 3 𝑋10−6
(0.1) 2 = 27 X 105 N/C
EB = 1
4𝜋∈𝑜
𝑞𝐵
𝑂𝐵2 = 9 𝑋 109 𝑋 3 𝑋10−6
(0.1) 2 = 27 X 105 N/C
ER = EA + EB = 27 x105 + 27 x 105 = 54 X 105 N/C
ER = 𝐹
𝑞
F = qER = 1.5 X 10-9 X 54 X 105
F = 8.1 X 10-3 N
2) Three charges each equal to +4nc are placed at the thrice corners of a
square of side 2cm find the electric field at the fourth corner
BD2 = AB2 + AD2
= 22 + 22
= 4 + 4 =8 = 2√2 cm
E = 1
4𝜋∈𝑜
𝑞
𝑟2
EA = 1
4𝜋∈𝑜
𝑞𝐴
𝐴𝐷2 = 9 𝑋 109 4 𝑋10−9
(2𝑋10−2) 2 = 9 X 104 N/C along AD
EC = 1
4𝜋∈𝑜
𝑞𝐶
𝐶𝐷2 = 9 𝑋 109 4 𝑋10−9
(2𝑋10−2) 2 = 9 X 104 N/C along CD
EB = 1
4𝜋∈𝑜
𝑞𝐵
𝐵𝐷2 = 9 𝑋 109 4 𝑋10−9
(2𝑋√2𝑋10−2) 2 = 4.5 X 104 N/C along BD
Resultant of EA and EC at D , ED = √𝐸𝐴2 + 𝐸𝐶2 𝜃 =90°
= √(9𝑋104) 2 + (9𝑋104) 2 cos90° = 0
= √162 𝑋 108 = 12.72 x 104 N/C
Total field at O due to all the charges
ED +EB = 12.72 X 104 + 4.5 104
= 17.22 X 104 N/C
3) ABC is a equilateral triangle of side 0.1m charged of 3nc and -3nc are
placed at the vertices A and b respectively . calculate the resultant
electric intensity at a point C
E1 = 1
4𝜋∈𝑜
𝑞𝐴
𝐴𝐶2 = 9 𝑋 109 𝑋 3 𝑋10−9
(0.1) 2 =
27
0.01
E1 = 2700 N/C
E2 = 1
4𝜋∈𝑜
𝑞𝐵
(𝐵𝐶)2 = 9 𝑋 109 𝑋 3 𝑋10−9
(0.1) 2 =
27
0.01
E2 = 2700 N/C
ER = √𝐸12 + 𝐸2
2 + 2𝐸1𝐸2𝑐𝑜𝑠𝜃 𝜃 =120° cos 120° = −1
2
√(2700) 2 + (2700) 2 + 2(2700)𝑋(2700)𝑐𝑜𝑠120°
ER = √(2700) 2 + (2700) 2 + 2(2700) 2𝑋 (−1
2)
= √(2700) 2 + (2700) 2 − (2700) 2
= √(2700) 2
= ER = 2700 N/C
ATOMS
Bohr’s postulates
1. An electron in an atom could revolve in certain stable orbits without the
emission of radiant energy
2. An electron revolves around the nucleus only in those orbits for which the
angular momentum is some integral multiple of ℎ
2𝜋 where h is the plank
constant i.e L = 𝑛ℎ
2𝜋 where n= 1,2,3,4………
3. An electron make a transition from one of its specified non- radiating
orbits to another of lower energy . When it does so , a photon is emitted
having energy equal to energy difference between the initial and final
states
i.e hr = Ei - Ef where Ei and Ef are the energies of the initial and final states
and Ei > Ef
EXPRESSION FOR RADIUS OF NTH ORBIT OF HYDROGEN ATOM
m is the mass of electron
e → charge on electron
r → radius of nth orbit
z → number of photons
v → velocity of electron
𝑚𝑣2
𝑟 =
1
4𝜋∈𝑜
(𝑍𝑒)(𝑒)
𝑟2
𝑚𝑣2 = 𝑍𝑒2
4𝜋∈𝑜𝑟 →①
According to Bohr’s postulate
mvr = 𝑛ℎ
2𝜋
v = 𝑛ℎ
2𝜋𝑚𝑟 → ②
using ① and ②
m = (𝑛ℎ
2𝜋𝑚𝑟) 2 =
𝑍𝑒2
4𝜋∈𝑜𝑟
r = 𝑛2ℎ2𝜖𝑜
𝜋𝑚𝑍𝑒2 → ③
for hydrogen atom Z=1 from equ ③
r = 𝑛2ℎ2𝜖𝑜
𝜋𝑚𝑒2
EXPRESSION FOR TOTAL ENERGY OF ELECTRON OF HYDROGEN ATOM
m – mass of electron
e – charge on electron
r – radius of nth orbit
Z – Atomic number
Total energy E = K.E + P.E →①
𝑚𝑣2
𝑟 =
1
4𝜋∈𝑜
𝑍𝑒2
𝑟2
K.E = 1
2 𝑚𝑣2 =
1
2 (
1
4𝜋∈𝑜 𝑍𝑒2
𝑟 ) → ②
P.E = 1
4𝜋∈𝑜 𝑍𝑒 (−𝑒)
𝑟 = -
1
4𝜋∈𝑜 𝑍𝑒2
𝑟 → ③
Substitute ② & ③ in ①
E =1
2(
1
4𝜋∈𝑜 𝑍𝑒2
𝑟 ) - 1
4𝜋∈𝑜 𝑍𝑒2
𝑟
E =−𝑍𝑒2
8𝜋𝜖0𝑟 → ④
But radius of the electron, r= 𝑛2ℎ2𝜖0
𝜋𝑚𝑍𝑒2 → ⑤
Substitute ⑤ in ④
E= −𝑍𝑒2
8𝜋𝜀𝑜 (𝑛2ℎ2𝜀𝑜𝜋𝑚𝑍𝑒2 )
E= −𝑚𝑧2𝑒4
8𝜖 02 𝑛2ℎ2
EXPRESSION FOR THE WAVE NUMBER OF THE SPECTRAL LINE.
En1→energy of n1thorbit
En2 →energy of n2thorbit
ℎ𝛾=En2 –En1
ℎ𝛾 = −𝑚𝑧2𝑒4
8∈0 2𝑛2 2 ℎ2 - (−𝑚𝑧2𝑒4
8∈0 2𝑛2 1 ℎ2)
ℎ𝛾 = 𝑚𝑧2𝑒4
8∈0 2ℎ2 - (
1
𝑛1 2 −
1
𝑛2 2 )
𝛾 = 𝑚𝑧2𝑒4
8∈0 2ℎ2 - (
1
𝑛1 2 −
1
𝑛2 2 )
But 𝛾 = 𝑐
⋋ where c is speed of light ,⋋ → wavelength.
1
⋋ =
𝑚𝑧2𝑒4
8∈0 2ℎ3 𝐶 - (
1
𝑛1 2 −
1
𝑛2 2 )
𝛾 = 1
⋋ = R 𝑧2 (
1
𝑛1 2 −
1
𝑛2 2 ) where R =
𝑚𝑒𝑡
8∈0 2ℎ3 𝐶
For hydrogen atom, z=1, 𝛾 = R (1
𝑛1 2 −
1
𝑛2 2 )
Rydberg constant , R= 1.097x 107 m-1
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