Physics Class 12 Derivations PDF

(Physics) INDEX S.NO TOPIC PAGE NO. 1. Electric Field on Axial Line of an Electric Dipole 𝟏 2. Electric Field on Equatorial Line of Dipole...

(Physics) INDEX S.NO TOPIC PAGE NO. 1. Electric Field on Axial Line of an Electric Dipole 𝟏 2. Electric Field on Equatorial Line of Dipole 𝟏 3. Electric Dipole in Uniform Electric Field 𝟐 4. Electric Field Due to a Line Charge 𝟐 5. Electric Field Due to Infinite Charged Plane Sheet 𝟐−𝟑 6. Electric Field Due to Charged Spherical Shell 𝟑 7. Electric Potential at any point due to an Electric Dipole 𝟒 Potential Energy of an Electric Dipole, when Placed in Uniform 8. 𝟒 Electric Field 9. Parallel Plate Capacitor 𝟒−𝟓 Capacitance of Capacitor, when a Dielectric Slab Completely Fills 10. 𝟓 the Space Between Plates 11. Energy Stored in a Charged Capacitor 𝟓 12. Drift Velocity 𝟕 13. Relation between Drift Velocity and Electric Current 𝟕 14. Electric Current and Current Density 𝟖 15. Cells in Series 𝟖 16. Cells in Parallel 𝟖−𝟗 17. Wheat Stone Bridge 𝟗 18. Motion of a Charged particle inside a Uniform Magnetic Field 𝟏𝟎 19. Magnetic Field at a point on the Axis of a Loop 𝟏𝟏 Force between two infinitely Long Parallel Current Carrying 20. 𝟏𝟏 − 𝟏𝟐 Conductors 21. Torque on a Current Loop placed in a Magnetic Field 𝟏𝟐 22. Self-Induction (of a Long Solenoid ) 𝟏𝟑 23. Mutual Induction (of Two Long Solenoids) 𝟏𝟑 24. Mean or Average Value of A.C. 𝟏𝟒 25. Root Mean Square(rms) or Virtual Value of A.C. 𝟏𝟒 26. A.C. Through LCR-Series Circuit 𝟏𝟓 − 𝟏𝟔 27. Power of an A.C. Circuit 𝟏𝟔 28. Mirror Formula (for Concave spherical mirror) 𝟏𝟕 29. Linear Magnification(for Concave spherical mirror) 𝟏𝟕 30. Refraction at Convex Spherical Surface 𝟏𝟖 31. Lens Maker's Formula 𝟏𝟖 − 𝟏𝟗 32. Refraction through a Prism 𝟏𝟗 33.a Simple Microscope Magnifying power (Image is formed at D) 20 33.b Simple Microscope Magnifying power (Image is formed at infinity). 𝟐𝟎 34.a Compound Microscope Magnifying power (Image is formed at D). 𝟐𝟏 34.b Compound Magnifying power (Image is formed at infinity). 𝟐𝟏 35. Astronomical Telescope (Image is formed at infinity). 𝟐𝟏 APNI KAKSHA 0 (Physics) 36. Laws of Reflection on the basis Wave Theory 𝟐𝟑 37. Laws Refraction on the Basis of Wave Theory 𝟐𝟑 38. Conditions for Constructive and Destructive Interference 24 39. Distance of Closest Approach 𝟐𝟓 40. Bohr's Theory of Hydrogen Atom 𝟐𝟓 41. Nuclear Density 𝟐𝟔 42. Mass Defect 𝟐𝟔 43. Binding Energy 𝟐𝟔 APNI KAKSHA 0 (Physics) ELECTRIC FIELD AND CHARGES 1. Electric Field on Axial line of an Electric Then, resultant electric field at point P is given by Dipole [𝟑 𝐦𝐚𝐫𝐤𝐬, 𝐂𝐁𝐒𝐄 𝟐𝟎𝟏𝟕, 𝟏𝟗, 𝟐𝟎, 𝟐𝟏] Consider an electric dipole consisting of charges −q and +q, separated by a distance 2a and placed in free space. ⃗ at point P due to the dipole will The electric field E ⃗ A (due to −q be the resultant of the electric fields E at point A ) and ⃗EB (due to +q at the B) i.e. ⃗E = ⃗EA + ⃗EB ⃗E = ⃗EA + ⃗EB Let ∠MPN = ∠PBN = θ. ⃗ B | >|E Also, |E ⃗ A |. Also ∠NPL = ∠PAB = θ ̂ ̂ So, ⃗E = ⃗EA + ⃗EB = (EA cos θ + EB cos θ)(−i) ⃗E = ((E ⃗ B ) − (E ⃗ A ) ) (𝑖) 1 q 1 q ̂ ⃗E = ⃗EA + ⃗EB = (2EA cos θ)(−i) or ⃗E = ⋅ 2 − ⋅ 4πε0 (r − a) 4πε0 (r + a)2 1 q 2a ̂ = ⋅ 2 2 × 2 (−𝑖) 1 (r + a)2 − (r − a)2 4πε0 (r + a ) (r + a2 )1/2 = ⋅q 1 q(2a) 4πε0 (r 2 − a2 )2 ̂ or E= ⋅ 2 (−𝑖) 1 q(4ra) 4πε0 (r + a2 )3/2 ⃗E = ⋅ 2 ̂ (𝑖) 4πε0 (r − a2 )2 ̂ , So, Now ⃗P = q(2a)(𝑖) 1 P ∴ E= ⋅ 2 ̂ (−𝑖) ̂ , then. Now, ⃗P = q(2a)(𝑖) 4πε0 (r + a2 )3/2 1 2Pr In vector notation, ∴ ⃗E = ⋅ 2 ̂ (𝑖) 4π ε0 (r − a2 )2 ⃗ 1 P ⃗ =− E ⋅ 2 4πε0 (r + a2 )3/2 In vector notation, For dipole is of small length, a = −1 u υ f u f PA = −u (Object distance) υ υ υ−f = −1= PA′ = −υ(Image dist. ) and PF = −f (focal length) u f f In the equation (iii), substituting for PA, PA′ and PF, So, we have 𝐟− 𝛖 𝐦= −υ − (−f) −υ υ−f υ υ υ 𝐟 = or = or − 1 = −f −u f u f u 1 1 1 𝟏 𝟏 𝟏 or − = or + = f υ u 𝐮 𝛖 𝐟 The above relation between u, υ and f is called mirror formula. APNI KAKSHA 17 (Physics) RAY OPTICS & OPTICAL INSTRUMENTS 30. Refraction at Convex Spherical Surface Since angles β and γ are small, we have [𝟑 𝐦𝐚𝐫𝐤𝐬, 𝐂𝐁𝐒𝐄 𝟐𝟎𝟏𝟕, 𝟏𝟖, 𝟏𝟗, 𝟐𝟎, 𝟐𝟏, 𝟐𝟑] r = tan γ − tan β … (v) Let us consider a convex spherical refracting From right angled triangles ANC and ANI, we have surface with AN AN AN AN tan γ = ≈ and tan β = ≈ μ2 = refractive index of medium 2 & NC PC NI PI μ1 = Refractive index of medium 1, In the equation (v), substituting for tan β and tan γ Let P = pole, C = center of curvature and we have PC = Principal axis of the convex surface. AN AN r= − … (vi) PC PI When object lies in the rarer medium and image By Snell’s law formed is real. μ1 sin i = μ2 sin r O = Object. Draw AN as perpendicular and take Since the angles i and r are also small, the above angle α, β & γ respectively in triangles. equation becomes Let ∠AOP = α; ∠AIP = β and ∠ACP = γ. μ1 i = μ2 r In triangle AOC, we have From the equations (iv) and (vi), substituting the i=α+γ (exterior angle property) … (i) values of i and r, we have AN AN AN AN μ1 ( + ) = μ2 ( − ) PO PC PC PI μ1 μ1 μ2 μ2 or + = − PO PC PC PI μ1 μ2 μ2 − μ1 or + = … (vii) PO PI PC Since angles α, β and γ will be small. As such, these Applying new cartesian sign conventions: angles may be replaced by their tan α etc. PO = −u (object distance) Therefore, equation (i) may be written as PI = +v (image distance)and i = tan α + tan γ … (ii) PC = +R (Radius of curvature) From right angled triangles ANO and ANC, we have Therefore, the equation (vii) becomes AN AN μ1 μ2 μ2 − μ1 𝛍𝟐 𝛍𝟏 𝛍𝟐 − 𝛍𝟏 tan α = and tan γ = + = or − = NO NC −u +v +R 𝐯 𝐮 𝐑 In the equation (ii), substituting for tan α and tan α, 31. Lens Maker's Formula we have [𝟑 𝐦𝐚𝐫𝐤𝐬, 𝐂𝐁𝐒𝐄 𝟐𝟎𝟏𝟕, 𝟏𝟖, 𝟐𝟎, 𝟐𝟏, 𝟐𝟐] AN AN i= + … (iii) NO NC NO ≈ PO and NC ≈ PC Therefore, the equation (iii) becomes AN AN i= + … (iv) PO PC Now, from triangle ACI, γ = r + β (by exterior angle property) or r = γ − β Let us consider a convex lens with refracting surface with APNI KAKSHA 18 (Physics) RAY OPTICS & OPTICAL INSTRUMENTS μ2 = refractive index of outer medium & 32. Refraction through a Prism μ1 = Refractive index of lens , [𝟑 𝐦𝐚𝐫𝐤𝐬, 𝐂𝐁𝐒𝐄 𝟐𝟎𝟏𝟕, 𝟏𝟖, 𝟐𝟎, 𝟐𝟏, 𝟐𝟐] Suppose that O is a point object placed on the KTS = δ is called the angle of deviation. principal axis of the lens. The surface XP1 Y forms the real image I1 (assuming that material of the lens extends beyond the face XP1 Y as such). It can be obtained* that μ1 μ2 μ2 − μ1 + = … (i) P1 O P1 I1 P1 C1 Since the lens is thin, the point P1 lies very close to the optical centre C of the lens. Therefore, we may write Since ∠TQO = ∠NQP = i and ∠RQO = r1 , we have P1 O ≈ CO; P1 I1 ≈ CI1 and P1 C1 ≈ CC1 ∠TQR = i − r1 𝛍𝟏 𝛍𝟐 𝛍𝟐 −𝛍𝟏 So, + 𝐂𝐈 = … (ii) Also, 𝐂𝐎 𝟏 𝐂𝐂𝟏 The image formed by first refraction will act as ∠TRO = ∠NSE = e and ∠QRO = r2. Therefore, virtual object for 2nd surface refraction. ∠TRQ = e − r2 μ2 μ1 μ2 − μ1 − + = … (iii) in triangle TQR, by exterior angle property P2 I1 P2 I P2 C2 δ = ∠TQR + ∠TRQ = (i − r1 ) + (e − r2 ) Again P2 I1 ≈ CI1 , P2 I ≈ CI and P2 C2 ≈ CC2 or 𝛅 = (𝐢 + 𝐞) − (𝐫𝟏 + 𝐫𝟐 ) … (i) Therefore, eq (iii) may be written as 𝛍𝟐 𝛍𝟏 𝛍𝟐 − 𝛍𝟏 In triangle QRO, the sum of the angles is 180∘. − + = … (iv) 𝐂𝐈𝟏 𝐂𝐈 𝐂𝐂𝟐 Therefore, Adding the eq (ii) and (iv), we have r1 + r2 + ∠QOR = 180∘ … (ii) μ1 μ2 μ2 μ1 μ2 − μ1 μ2 − μ1 + − + = + In quadrilateral AQOR, CO CI1 CI1 CI CC1 CC2 A + ∠QOR = 180∘ … (iii) μ1 μ1 1 1 or + = (μ2 − μ1 ) ( + ) … (v) From the equations (ii) and (iii), we have CO CI CC1 CC2 Applying the new cartesian sign conventions: r1 + r2 = A … (iv) CO = −u (object distance) In the equation (i), substituting for (r1 + r2 ) we CI = +v (Final image distance) have 𝛅 = (𝐢 + 𝐞) − 𝐀 … (v) CC1 = +R1 and CC2 = −R 2 (Radii of curvature) Also, when δ = δm ; (in minimum deviation μ1 μ1 1 1 position), + = (μ2 − μ1 ) ( + ) −u +v +R1 −R 2 e = i and r2 = r1 = r = A/2 (say) Dividing both sides of the above equation by μ1 , we Also, setting δ = δm and e = i in the equation (v), have we have Since μ2 /μ1 = μ, we have A + δm = i + i or i = (A + δm )/2 𝟏 𝟏 𝟏 𝟏 − = (𝛍 − 𝟏) ( − ) … (vi) The refractive index of the material ( a μg or simply 𝐯 𝐮 𝐑𝟏 𝐑𝟐 Also if u = CF1 = −f1 (focal length ), then v = ∞ μ) of the prism is given by Setting the above condition in the equation (vi), we sin i μ = have sin r 1 1 1 1 𝐬𝐢𝐧(𝐀 + 𝛅𝐦 )/𝟐 − + = (μ − 1) ( − ) ∴ 𝛍= −f1 ∞ R1 R 2 𝐬𝐢𝐧 𝐀/𝟐 𝟏 𝟏 𝟏 = (𝛍 − 𝟏) ( − ) … (ix) 𝐟 𝐑𝟏 𝐑𝟐 APNI KAKSHA 19 (Physics) RAY OPTICS & OPTICAL INSTRUMENTS 33. Simple Microscope (Magnifying Glass) Now, CA = u and CA′ = D [𝟑 𝐦𝐚𝐫𝐤𝐬, 𝐂𝐁𝐒𝐄 𝟐𝟎𝟏𝟖, 𝟐𝟐, 𝟐𝟑] Therefore, the equation (ii) becomes A convex lens of short focal length can be used to D M= … (iii) see magnified image of a small object and is called a u magnifying glass or a simple microscope. u = −u or υ = −D So, 33.a - Magnifying power- When image is formed Magnifying power of simple microscope at D (least distance of distinct vision=25cm). 𝐀𝐧𝐠𝐥𝐞 𝐦𝐚𝐝𝐞 𝐛𝐲 𝐢𝐦𝐚𝐠𝐞 𝐨𝐧 𝐞𝐲𝐞 ( 𝐰𝐡𝐞𝐧 𝐤𝐞𝐩𝐭 𝐚𝐭 𝐃) Therefore, the above equation becomes by lens = 𝐀𝐧𝐠𝐥𝐞 𝐦𝐚𝐝𝐞 𝐛𝐲 𝐨𝐛𝐣𝐞𝐜𝐭 𝐨𝐧 𝐞𝐲𝐞 (𝐰𝐡𝐞𝐧 𝐤𝐞𝐩𝐭 𝐚𝐭 𝐃) formula 1 1 1 1 1 1 Let ∠A′ CB ′ = β be the angle subtended by the − + = or − = −u −D f u D f image at the eye. Cut A′ Q equal to AB(object size) D D D D or − 1 = or = 1 + … (iv) and join QC Then, ∠A′ CQ′ = α is the angle u f u f subtended by the object at the eye, when it is placed From the equations (iii) and (iv), we have at the least distance of distinct vision. 𝐃 𝐌=𝟏+ … (v) By definition, the magnifying power of the simple 𝐟 microscope is given by 33.b - Magnifying power (When image is formed β at infinity). M= α u = −f and CA′ = −D (Note: Remember the first clear image is always seen at D only) In practice, the angles α and β are small. Therefore, Therefore, the above equation gives the angles α and β can be replaced by their tangents D D M= = … (vii) i.e. u f tan β 34. Compound Microscope M= … (i) tan α [𝟑 𝐦𝐚𝐫𝐤𝐬, 𝐂𝐁𝐒𝐄 𝟐𝟎𝟏𝟖, 𝟐𝟎, 𝟐𝟏] From the right angled △ CA′ Q, A′ Q AB tan α = ′ = ′ (∵ A′ Q = AB) CA CA Also, from the right angled △ ABC AB tan β = CA Substituting for tan α and tan β in the equation we have AB/CA CA′ M= or M = … (ii) AB/CA′ CA APNI KAKSHA 20 18 10 (Physics) RAY OPTICS & OPTICAL INSTRUMENTS For a compound microscope two lenses, eyepiece υe = −D and fe = +fe of focal lenght (fe )and objective of focal lenght (f0 ) In the above equation, substituting for υe and fe , we are used to achieve greater magnification then have simple microscope. First clear image is formed at D D me = 1 + … (iv) fe (least distance of distinct vision = 25cm) So, putting values in M = mo × me we get So, 𝛖𝐨 𝐃 Magnifying power of Compound microscope 𝐌= (𝟏 + ) … (vii) 𝐮𝐨 𝐟𝐞 𝐀𝐧𝐠𝐥𝐞 𝐦𝐚𝐝𝐞 𝐛𝐲 𝐢𝐦𝐚𝐠𝐞 𝐨𝐧 𝐞𝐲𝐞 ( 𝐰𝐡𝐞𝐧 𝐤𝐞𝐩𝐭 𝐚𝐭 𝐃) = 34.b - Magnifying power - When image is formed 𝐀𝐧𝐠𝐥𝐞 𝐦𝐚𝐝𝐞 𝐛𝐲 𝐨𝐛𝐣𝐞𝐜𝐭 𝐨𝐧 𝐞𝐲𝐞 (𝐰𝐡𝐞𝐧 𝐤𝐞𝐩𝐭 𝐚𝐭 𝐃) Let ∠A′′ C′B ′′ = β be the angle subtended by the at infinity. υ υ image at the eye. extend A′′ Q equal to We know , M = mo × me & m0 = u0 & me = ue, 0 e AB(object size) and join QC Then, ∠A′ ′C′Q = α is For image at infinity 𝒖𝒆 = −𝒇𝒆 & 𝝊𝒆 = −𝑫 the angle subtended by the object at the eye, when Here fe is the focal length of the eye lens. it is placed at the least distance of distinct vision. (Note: Remember the first clear image is always seen By definition, the magnifying power of the simple at D only so 𝜐𝑒 = −𝐷) microscope is given by 𝛖𝐨 𝐃 So, 𝐌 = ×𝐟 … (ix) β 𝐮𝟎 𝐞 M= α 35. Astronomical Telescope (Refracting Type) Since the angles α and β are small, they can be [𝟑 𝐦𝐚𝐫𝐤𝐬, 𝐂𝐁𝐒𝐄 𝟐𝟎𝟏𝟕, 𝟏𝟖, 𝟐𝟏] replaced by their tangents i.e. An astronomical telescope is used to see the tan β M= … (i) heavenly objects. tan α A′′ Q AB An astronomical telescope consists of two lens Also tan α = = ′ ′′ (∵ A′′ Q = AB) C′ A′′ ′′ ′′ CA systems. The lens system facing the object is called A B Also, tan β = ′ ′′ objective. It has large aperture and is of large focal CA Multiplying and dividing by A′ B ′ , we have length (f0 ). The other lens system is called eye- A′′ B ′′ A′ B ′ A′ B ′ A′′ B ′′ piece. It has small aperture and is of short focal M= × ′ ′= × ′ ′ AB AB AB AB length (fe ). Also the first clear image is formed at D A′ B ′ Also υ = uo = m0 = magnification of object lens (least distance of distinct vision = 25cm) AB o Magnifying power of refracting telescope ( υo & uo = Image & object dist. from object lens)& A′′ B′′ υe 𝐀𝐧𝐠𝐥𝐞 𝐦𝐚𝐝𝐞 𝐛𝐲 𝐢𝐦𝐚𝐠𝐞 𝐨𝐧 𝐞𝐲𝐞 (𝐰𝐡𝐞𝐧 𝐤𝐞𝐩𝐭 𝐚𝐭 𝐃) = = me = magnification of eye lens = A′ B ′ ue 𝐀𝐜𝐭𝐮𝐚𝐥 𝐚𝐧𝐠𝐥𝐞 𝐨𝐟 𝐨𝐛𝐣𝐞𝐜𝐭 𝐨𝐧 𝐞𝐲𝐞 (υe & ue = Image & object distance from eye lens) So, 𝐌 = 𝐦𝐨 × 𝐦𝐞 … (ii) 34.a - Magnifying power - When image is formed at D(least distance of distinct vision). Now, for the eye lens, the lens equation may be written as 1 1 1 υe υe Thus, ∠A′ CB ′ = α may be considered as the angle − + = or =1− ue υe fe ue fe subtended by object at the eye. So putting value for me we have, Let ∠A′ C′ B ′ = β. Then, by definition, υe me = 1 − … … … … (iii) β fe M= α Applying the new Cartesian sign conventions: Since the angles α and β are small, APNI KAKSHA 21 (Physics) RAY OPTICS & OPTICAL INSTRUMENTS α ≈ tan α and β ≈ tan β tan β In the equation (i), substituting for tan α and tan β, ∴M= … (i) tan α we have A′ B ′ From the right angled △ CA′ B ′ , tan α = CA′ C ′ A′ A′ B ′ = and from the right angled ΔC ′ A′ B ′ , tan α = ′ ′ B ′ /CA′ CA In the equation (i), substituting for tan α and tan β, we have S ′ ′ ′ ′ ′ A B /C A CA M= ′ ′ ′ = ′ ′ … (ii) A B /CA CA Magnifying power - When image is formed at infinity. (Note: Remember the first clear image is always seen at D only) Applying the new cartesian sign conventions: CA′ = +fo and C ′ A′ = −fℯ Substituting for CA′ and C ′ A′ in the equation (ii), we have 𝐟𝐨 𝐌=− 𝐟𝐞 APNI KAKSHA 22 (Physics) WAVE OPTICS 36. Laws of Reflection on Wave Theory Let ML be a beam of light that refracts to second [𝟑 𝐦𝐚𝐫𝐤𝐬, 𝐂𝐁𝐒𝐄 𝟐𝟎𝟏𝟕, 𝟏𝟖, 𝟐𝟎] medium from XY boundary. Let ML be a beam of light that reflects back from Here PA is wave front for incident beam and P’A’ is surface XY. wave front for refracted beam. Here PA is wave front for incident beam and P’A’ is Let time taken (t) by light to go from the point P to wave front for reflected beam. If c is velocity of P′ and in same time let A reaches to A’ after light, then time taken (t) by light to go from the refraction. point P to P′ and by light to go from A to A’ will be ∠LAN = i , ∠N ′ AA′ = r (angle of refraction) same as both lie on wave fronts. By using properties of complementary angle ∠PAP′ = i , ∠AP′A′ = r We have PP′ AA′ sin i = & sin r = ……… (i) AP′ AP′ Here PP’ = ct and AA’ =c’t c = speed of light in air c’ = speed of light in denser medium c ∠LAN = i , ∠NAA′ = r (angle of reflection) μ= = refractive index of denser medium ……. (ii) c′ By using properties of complementary angle so, from equation (i) & (ii) ∠PAP′ = i , ∠AP′A′ = r sin i PP′ ct c = = = =μ We have sin r AA′ c′t c′ PP′ AA′ 𝐬𝐢𝐧 𝐢 sin i = AP′ & sin r = AP′ ……… (i) = 𝛍 𝐬𝐢𝐧 𝐫 Here PP’ = ct and AA’ =ct Hence, the laws of refraction (Snell’s law ) is proved putting the values in (i) on the basis of the wave theory. ct ct sin i = & sin r = AP′ AP′ so, sin i = sin r 38. Conditions for Constructive and Destructive or i = r Interference i.e. the angle of incidence is equal to the angle of [𝟑 𝐦𝐚𝐫𝐤𝐬, 𝐂𝐁𝐒𝐄 𝟐𝟎𝟏𝟕, 𝟐𝟎, 𝟐𝟐] reflection. (laws of reflection ) Let a source of monochromatic light S illuminates 37. Refraction on The Basis of Wave Theory two narrow slits S1 and S2. The two illuminated slits [𝟑 𝐦𝐚𝐫𝐤𝐬, 𝐂𝐁𝐒𝐄 𝟐𝟎𝟏𝟕, 𝟏𝟗, 𝟐𝟏] act as the two coherent sources. At the centre O of the screen, the intensity of light is maximum and it is called central maximum. Condition for maximum and minimum. Let the displacements of the waves from the sources S1 and S2 at point P on the screen at any time t be given by y1 = a1 sin ωt and y2 = a2 sin(ωt + ϕ), where ϕ is the constant phase difference between the two waves. APNI KAKSHA 23 (Physics) WAVE OPTICS So superimposed wave will be For Destructive interference. y = y1 + y2 = a1 sin ωt + a2 sin(ωt + ϕ) From equation (iv) it follows that the intensity of y = (a1 + a2 cos ϕ) sin ωt + a2 sin ϕ cos ωt … (i) light at point P will be minimum, if Let a1 + a2 cos ϕ = Acos θ … (ii) cos ϕ = −1 or ϕ = π, 3π, 5π, …. and a2 sin ϕ = Asin θ … (iii) or ϕ = (2n + 1) π, Then, the equation (i) becomes where n = 0,1,2, … y = Acos θ sin ωt + Asin θ cos ωt Also, from the equations (vi) and (viii), we have or y = Asin(ωt + θ) 2π x = (2n + 1)π Also Squaring and adding both sides of the λ 𝛌 equations (ii) and (iii), we obtain or 𝐱 = (𝟐𝐧 + 𝟏) 𝟐 where n = 0,1,2 ….. n A2 cos 2 θ + A2 sin2 θ = (a1 + a2 cos ϕ)2 + a22 sin2 ϕ or A2 = a21 + a22 , (cos2 ϕ + sin2 ϕ) + 2a1 a2 cos ϕ or A2 = a21 + a22 + 2a1 a2 cos ϕ ….. (iv) For constructive interference the intensity of light will be maximum so , A=max So, cos ϕ = 1 𝟐𝛑 So , 𝛟= 𝐱 = 𝟐𝐧𝛑 , or 𝐱 = 𝐧𝛌 where 𝛌 n = 0,1,2,3….. n APNI KAKSHA 24 22 (Physics) ATOMS 39. Distance of Closest Approach If m and v are mass and orbital velocity of the The value of the distance of closest approach gives electron, then the centripetal force required by the an estimate of the size of the nucleus. electron to move in circular orbit of radius r is given mv2 by Fc = r … (ii) The electrostatic force of attraction (Fe ) between the electron and the nucleus provides the necessary centripetal force (Fc ) to the electron. Therefore, from the equations (i) and (ii), we have mv 2 1 e2 1 e2 = ⋅ 2 or mv 2 = ⋅ … (iii) r 4πε0 r 4πε0 r Consider that an α-particle of mass m possesses According to Bohr's quantization condition, angular initial velocity u, when it is at a large distance from momentum of the electron, the nucleus of an atom having atomic number Z. At h nh the distance of closest approach, the kinetic energy mvr = n or v = … (iv) 2π 2πmr of the α-particle is completely converted into In the equation (iii), putting the value of v, we have potential energy. Mathematically, nh 2 1 e2 1 1 2e(Ze) m( ) = ⋅ mu2 = ⋅ 2πmr 4πε0 r 2 4πε0 r0 𝐧 𝟐 𝐡𝟐 1 2Ze2 or 𝐫 = 𝟒𝛑𝛆𝟎 ⋅ … (v) ∴ r0 = ⋅ … (i) 𝟒𝛑𝟐 𝐦𝐞𝟐 4πε0 1 mu2 2 Since n = 1,2,3,4 … , The equation (i) is the expression for the distance Also of closest approach. Energy of the electron in nth orbit of a hydrogen-like In Geiger-Marsden experiment, α-particles of atom is given by kinetic energy 5.5 MeV were directed towards the 1 2 2π2 Z2 me4 En = − ( ) ⋅ gold nucleus (Z = 79). By calculating the distance 4πε0 n2 h2 of closest approach r0 , an estimate of the size of the where Z is atomic number of the atom. nucleus can be made. The calculations show that r0 nh 1 4π2 me2 v= ( ⋅ ) comes out to be 4 ⋅ 13 × 10−14 m. Thus, size of the 2πm 4πε0 n2 h2 nucleus is of the order of 10−14 m. 1 2πe2 or v = ⋅ … (vi) 40. Bohr's Theory of Hydrogen Atom 4πε0 nh [𝟑 𝐦𝐚𝐫𝐤𝐬, 𝐂𝐁𝐒𝐄 𝟐𝟎𝟏𝟕, 𝟏𝟖, 𝟐𝟎, 𝟐𝟏, 𝟐𝟐] 1 Obviously, Ek = mv 2 In a hydrogen atom, an electron having charge −e 2 revolves round the nucleus having charge +e in a Using the equation (iii), we have circular orbit of radius r as shown in Fig. 1 e2 Ek = ⋅ 4πε0 2r 1 (+e)(−e) 1 e2 Ep = ⋅ =− ⋅ 4πε0 r 4πε0 r The total energy of electron revolving round the nucleus in the orbit of radius r is given by 1 e2 1 e2 E = Ek + Ep = ⋅ + (− ⋅ ) 4πε0 2r 4πε0 r The electrostatic force of attraction between the 1 e2 or E=− ⋅ nucleus and the electron is given by 4πε0 2r 1 e×e 1 e2 𝟏 𝟐 𝟐𝛑𝟐 𝐦𝐞𝟒 Fe = ⋅ 2 = ⋅ … (i) or 𝐄𝐧 = − ( ) ⋅ 4πε0 r 4πε0 r 2 𝟒𝛑𝛆𝟎 𝐧𝟐 𝐡𝟐 APNI KAKSHA 25 (Physics) NUCLEI 41. Nuclear Density The mass defect can also be expressed in another Let ρ be the density of the nucleus of an atom, form as explained below: whose mass number is A. Adding and subtracting the mass of Z electrons i.e. mass of the nucleus of the atom of mass number A Zme on the R.H.S. of equation (i), we have = A a.m.u. = A × 1.660565 × 10−27 kg Δm = [Zmp + (A − Z)mn + Zme ] − mN ( Z X A ) − Zme If R is the radius of the nucleus, then = [Z(mp + me ) + (A − Z)mn ] − [mN ( Z X A ) + Zme ] 4 4 1 3 Now, mp + me = m( 1 H1 ), mass of hydrogen atom volume of nucleus = πR3 = π (R 0 A3 ) 3 3 & mN ( Z X A ) + Zme = m( Z X A ), mass of the atom 4 = πR 0 3 A A 3 ZX Taking R 0 = 1 ⋅ 1 × 10−15 m, we have Therefore, 4 𝚫𝐦 = [𝐙𝐦( 𝟏 𝐇 𝟏 ) + (𝐀 − 𝐙)𝐦𝐧 ] − 𝐦( 𝐙 𝐗 𝐀 ) … (ii) volume of the nucleus = π(1 ⋅ 1 × 10−15 )3 × Am3 3 mass of nucleus 43. Binding energy. Density of the nucleus, ρ = volume of nucleus [𝟐 𝐦𝐚𝐫𝐤𝐬, 𝐂𝐁𝐒𝐄 𝟐𝟎𝟏𝟕, 𝟏𝟗, 𝟐𝟐, 𝟐𝟑] A × 1 ⋅ 660565 × 10−27 Thus, the binding energy of a nucleus may be = 4 −15 )3 × A defined as the energy equivalent to the mass defect 3 π(1 ⋅ 1 × 10 of the nucleus. = 𝟐 ⋅ 𝟗𝟕 × 𝟏𝟎𝟏𝟕 𝐤𝐠 𝐦−𝟑 (Independent of A) If Δm is mass defect of a nucleus, then according to 42. Mass Defect Einstein's massenergy relation, binding energy of [𝟐 𝐦𝐚𝐫𝐤𝐬, 𝐂𝐁𝐒𝐄 𝟐𝟎𝟏𝟕, 𝟏𝟖, 𝟐𝟎, 𝟐𝟐, 𝟐𝟑] the nucleus = Δmc 2 (in joule) The difference between the sum of the masses of Binding energy = [{𝐙𝐦𝐩 + (𝐀 − 𝐙)𝐦𝐧 } the nucleons constituting a nucleus and the rest mass of the nucleus is known as mass defect. It is − 𝐦𝐍 ( 𝐙 𝐗 𝐀 )] × 𝐜 𝟐 denoted by Δm. Here, Let us calculate the mass defect in case of the mN ( Z X A ) is mass of the nucleus of the atom Z X A. nucleus of an atom ZX A. The nucleus of such an mp = mass of proton , atom contains Z protons and (A-Z) neutrons. mn = mass of neutron Therefore, A = Mass number , Z = Atomic number mass of the nucleons = Zmp + (A − Z)mn 1amu × 𝑐 2 =931.5 Mev (All masses are kept in amu) If mN ( Z X A ) is mass of the nucleus of the atom Z X A, The mass defect can also be expressed in another then the mass defect is given by form: 𝚫𝐦 = [𝐙𝐦𝐩 + (𝐀 − 𝐙)𝐦𝐧 ] − 𝐦𝐍 ( 𝐙 𝐗 𝐀 ) … (i) Δm = [Zm( 1 H1 ) + (A − Z)mn ] − m( 𝐙 X A ) Here, Binding energy = [{𝐙𝐦( 𝟏 𝐇𝟏 ) + (𝐀 − 𝐙)𝐦𝐧 } mN ( Z X A ) is mass of the nucleus of the atom Z X A. − 𝐦( 𝐙 𝐗 𝐀 )] × 𝐜 𝟐 mp = mass of proton , Here m( 1 H1 ) = mass of hydrogen atom mn = mass of neutron m( Z X A ) = mass of the atom Z X A A = Mass number , Z = Atomic number APNI KAKSHA 26