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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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