CHAPTER – 23 HEAT AND TEMPERATURE EXERCISES Ice point = 20° (L0) L1 = 32° Steam point = 80° (L100) L1  L 32  20  100 = T=  100 = 20°C L100  L 80  20 Ptr = 1.500 × 10 Pa P = 2.050 × 10 Pa We know, For constant volume gas Thermometer 2.050  10 P  273.16 K =  273.16 = 373.31 Ptr 1.500  10 Pressure Measured at M.P = 2.2 × Pressure at Triple Point 2.2  Ptr P  273.16 =  273.16 = 600.952 K  601 K T= Ptr Ptr Ptr = 40 × 10 Pa, P = ? T= T = 100°C = 373 K, T= P  273 16 K Ptr T  Ptr 373  49  10 3 = = 54620 Pa = 5.42 × 10 pa ≈ 55 K Pa 273.16 273.16 P1 = 70 K Pa, P2 = ? T2 = 373K T1 = 273 K, P= T= T2 = P1  273.16 Ptr P2  273.16 Ptr  273 =  373 = 70  10  273.16 Ptr P2  273 70  273.16  10 Pice point = P0° = 80 cm of Hg Psteam point = P100° 90 cm of Hg P0 = 100 cm P  P0 80  100  100 =  100 = 200°C t= 90  100 P100  P0 V T0 T0 = 273, V  V V = 1800 CC, V = 200 CC 1800  273 = 307.125  307 T = 1600 Rt = 86; R0° = 80; R100° = 90 R t  R0 86  80 t=  100 =  100 = 60°C R100  R 90  80 T = R at ice point (R0) = 20 R at steam point (R100) = 27.5 R at Zinc point (R420) = 50 R = R0 (1+  +  )  R100 = R0 + R0  +R0   R0 R  100 =  +  R0 23.1  Ptr 70  273.16  10 273  P2 = 373  70  10 = 95.6 K Pa 273 23.Heat and Temperature 27.5  20 =  × 100 +  × 10000 20  = 100  + 10000  20 50  R 2 R420 = R0 (1+  +  )  =  +  R0  50  20 = 420 ×  + 176400 ×     420  + 176400  20  = 100  + 10000     420  + 176400  20 –5 L1 = ?, L0 = 10 m,  = × 10 /°C, t= 35 –5 –4 L1 = L0 (1 + t) = 10(1 + 10 × 35) = 10 + 35 × 10 = 10.0035m t1 = 20°C, t2 = 10°C, L1 = 1cm = 0.01 m, L2 =? –5 steel = 1.1 × 10 /°C –5 –4 L2 = L1 (1 + steelT) = 0.01(1 + 101 × 10 × 10) = 0.01 + 0.01 × 1.1 × 10 –6 –6 –6 = 10 × 10 + 1.1 × 10 = 10 (10000 + 1.1) = 10001.1 –2 =1.00011 × 10 m = 1.00011 cm –5  = 11 × 10 /°C L0 = 12 cm, tw = 18°C ts = 48°C –5 Lw = L0(1 + tw) = 12 (1 + 11 × 10 × 18) = 12.002376 m –5 Ls = L0 (1 + ts) = 12 (1 + 11 × 10 × 48) = 12.006336 m L12.006336 – 12.002376 = 0.00396 m  0.4cm –2 d1 = cm = × 10 t1 = 0°C, t2 = 100°C –5 al = 2.3 × 10 /°C –2 –5 d2 = d1 (1 + t) = × 10 (1 + 2.3 × 10 10 ) = 0.02 + 0.000046 = 0.020046 m = 2.0046 cm –5 Lst = LAl at 20°C Al = 2.3 × 10 /°C –5 st = 1.1 × 10 /°C So, Lost (1 – st × 20) = LoAl (1 – AI × 20)  10 11 12 13 14 (a)  Lo st (1   Al  20)  2.3  10 5  20 0.99954 = = = = 0.999 Lo Al (1   st  20) 0.99978  1.1 10   20 (b)  Lo 40st (1   AI  40)  2.3  10 5  20 0.99954 = = = = 0.999 Lo 40 Al (1   st  40) 0.99978  1.1 10   20 = Lo Al  2.3  10 5  10 0.99977  1.00092  = = 1.0002496 ≈1.00025 Lo st 273 1.00044 Lo100 Al (1   Al  100 ) 0.99977  1.00092 = = = 1.00096 Lo100St (1   st  100 ) 1.00011 15 (a) Length at 16°C = L L=? T1 =16°C, –5  = 1.1 × 10 /°C –5 L = L = L × 1.1 × 10 × 30 T2 = 46°C    L  L % of error =   100 % =   100 % = 1.1 × 10–5 × 30 × 100% = 0.033%    L  (b) T2 = 6°C  L  L   –5  100 % =   100 % = – 1.1 × 10 × 10 × 100 = – 0.011% % of error =   L  L   23.2 23.Heat and Temperature –3 16 T1 = 20°C, L = 0.055mm = 0.55 × 10 m –6 st = 11 × 10 /°C t2 = ? We know, L = L0T In our case, –3 –6 0.055 × 10 = × 1.1 I 10 × (T1 +T2) –3 –3 0.055 = 11 × 10 × 20 ± 11 × 10 × T2 T2 = 20 + = 25°C or 20 – = 15°C The expt Can be performed from 15 to 25°C 3 ƒ4°C = g/m 17 ƒ0°C=0.098 g/m , ƒ C 1 ƒ0°C =  0.998 =  + 4 =  T 1   0.998    = 0.0005 ≈ × 10–4 0.998 -4 As density decreases  = –5 × 10 18 Iron rod Aluminium rod LAl LFe –8 –8 Fe = 12 × 10 /°C Al = 23 × 10 /°C Since the difference in length is independent of temp Hence the different always remains constant …(1) LFe = LFe(1 + Fe × T) LAl = LAl(1 + Al × T) …(2) LFe – LAl = LFe – LAl + LFe × Fe × T – LAl × Al × T L Fe  23 = Al = = 23 : 12 L Al  Fe 12 4+= 2 19 g1 = 9.8 m/s , T1 = 2 l1 T2 = 2 g1 Steel = 12 × 10 T1 = 20°C T1 = T2  2 g2 = 9.788 m/s l1 g1 –6 = 2 l2 g2 = 2 l1(1  T ) g /°C T2 = ? l1(1  T ) g2  12  10 6  T = 9.788 9.788 –6   = 12 × 10 T   l1 l (1  T ) = g1 g2 9.788 –6 = 1+ 12 × 10 × T 0.00122  T = 12  10   T2 = – 101.6 + 20 = – 81.6 ≈ – 82°C    T2 – 20 = – 101.6 20 Given dAl = 2.000 cm dSt = 2.005 cm, –6 –6 S = 11 × 10 /°C Al = 23 × 10 /°C ds = 2.005 (1+ s T) (where T is change in temp.) –6  ds = 2.005 + 2.005 × 11 × 10 T –6 dAl = 2(1+ Al T) = + × 23 × 10 T The two will slip i.e the steel ball with fall when both the diameters become equal So, –6 –6  2.005 + 2.005 × 11 × 10 T = + × 23 × 10 T -6  (46 – 22.055)10 × T = 0.005  T = 0.005  10 = 208.81 23.945 23.3 Steel Aluminium 23.Heat and Temperature Now T = T2 –T1 = T2 –10°C [ T1 = 10°C given] T2 = T + T1 = 208.81 + 10 = 281.81 21 The final length of aluminium should be equal to final length of glass Let the initial length o faluminium = l l(1 – AlT) = 20(1 – 0) –6 –6  l(1 – 24 × 10 × 40) = 20 (1 – × 10 × 40)  l(1 – 0.00096) = 20 (1 – 0.00036) 20  0.99964 l= = 20.012 cm 0.99904 Let initial breadth of aluminium = b b(1 – AlT) = 30(1 – 0) b = 30  (1   10 6  40) (1  24  10 22 Vg = 1000 CC, VHg = ? 6  40) = 30  0.99964 = 30.018 cm 0.99904 T1 = 20°C –4 Hg = 1.8 × 10 /°C –6 g = × 10 /°C T remains constant Volume of remaining space = Vg – VHg Now Vg = Vg(1 + gT) …(1) VHg = VHg(1 + HgT) …(2) Subtracting (2) from (1) Vg – VHg = Vg – VHg + VggT – VHgHgT Vg  Hg 1.8  10 4 1000  =  = VHg g VHg  10   VHG =  10 3 = 500 CC. 1.8  10  23 Volume of water = 500cm Area of cross section of can = 125 m Final Volume of water –4 = 500(1 + ) = 500[1 + 3.2 × 10 × (80 – 10)] = 511.2 cm The aluminium vessel expands in its length only so area expansion of base cab be neglected Increase in volume of water = 11.2 cm Considering a cylinder of volume = 11.2 cm 11.2 Height of water increased = = 0.089 cm 125 24 V0 = 10 × 10× 10 = 1000 CC T = 10°C, VHG – Vg = 1.6 cm –6 –6 g = 6.5 × 10 /°C, Hg = ?, g= × 6.5 × 10 /°C …(1) VHg = vHG(1 + HgT) Vg = vg(1 + gT) …(2) VHg – Vg = VHg –Vg + VHgHg T – Vgg T –6  1.6 = 1000 × Hg × 10 – 1000 × 6.5 × × 10 × 10 1.6  6.3   10 2 –4 –4 = 1.789 × 10  1.8 × 10 /°C 10000 3 25 ƒ = 880 Kg/m , ƒb = 900 Kg/m –3 T1 = 0°C,  = 1.2 × 10 /°C, –3 b = 1.5 × 10 /°C The sphere begins t sink when, (mg)sphere = displaced water  Hg = 23.4 23.Heat and Temperature  Vƒ g = Vƒb g ƒb ƒ          b  880 900 =  1.2  10 3   1.5  10 3  –3 –3  880 + 880 × 1.5 × 10 () = 900 + 900 × 1.2 × 10 () –3 –3  (880 × 1.5 × 10 – 900 × 1.2 × 10 ) () = 20 –3  (1320 – 1080) × 10 () = 20   = 83.3°C ≈ 83°C L = 100°C A longitudinal strain develops if and only if, there is an opposition to the expansion Since there is no opposition in this case, hence the longitudinal stain here = Zero 1 = 20°C, 2 = 50°C –5 steel = 1.2 × 10 /°C Longitudinal stain = ? L L Stain = = =  L L –5 –4 = 1.2 × 10 × (50 – 20) = 3.6 × 10 –6 A = 0.5mm = 0.5 × 10 m T1 = 20°C, T2 = 0°C –5 11 s = 1.2 × 10 /°C, Y = × × 10 N/m Decrease in length due to compression = L …(1) Stress F L FL Y=  =  L = …(2) Strain A L AY Tension is developed due to (1) & (2) Equating them, FL  F = AY L = AY -5 –5 11 = 1.2 × 10 × (20 – 0) × 0.5 × 10 × 10 = 24 N 1 = 20°C, 2 = 100°C –6 A = 2mm = × 10 m –6 11 steel = 12 × 10 /°C, Ysteel = × 10 N/m Force exerted on the clamps = ?  26 27 28 29 F    A  = Y  F = Y  L  L = YLA = YA L Strain L 11 –6 –6 = × 10 × × 10 × 12 × 10 × 80 = 384 N 30 Let the final length of the system at system of temp 0°C = ℓ Initial length of the system = ℓ0 When temp changes by   Strain of the system =    total stress of system But the total strain of the system = total young' s mod ulusof of system Now, total stress = Stress due to two steel rod + Stress due to Aluminium = ss + s ds  + al at  = 2% s  + 2 Aℓ  Now young’ modulus of system = s + s + al = 2s + al 23.5  1m Steel Aluminium Steel 23.Heat and Temperature  Strain of system =  2 s  s    s  al   s   al   0 2 s  s    s  al  = 0  s   al 1   al  al  2 s  s    ℓ = ℓ0    al  2 s   31 The ball tries to expand its volume But it is kept in the same volume So it is kept at a constant volume So the stress arises P V =BP= B = B × V  V     v  11 –6 –6 = B × 3 = 1.6 × 10 × 10 × × 12 × 10 × (120 – 20) = 57.6 × 19  5.8 × 10 pa  32 Given 0 = Moment of Inertia at 0°C  = Coefficient of linear expansion To prove,  = 0 = (1 + 2) Let the temp change to  from 0°C T =  Let ‘R’ be the radius of Gyration, 0 = MR where M is the mass Now, R = R (1 + ), 2 2 Now,  = MR = MR (1 + )  = MR (1 + 2) 2 [By binomial expansion or neglecting   which given a very small value.] (proved) So,  = 0 (1 + 2) 33 Let the initial m. at 0°C be 0  K  = 0 (1 + 2) T = 2 (from above question) At 5°C, T1 = 2  (1  2) = 2 K At 45°C, T2 = 2  (1  90 )  (1  2 45) = 2 K K T2 = T1  90 =  10  90  2.4  10 5  10  2.4  10 5  (1  25)  (1  10 ) = 2 K K 1.00216 1.00024  T –2 % change =   1  100 = 0.0959% = 9.6 × 10 % T   T2 = 50°C, T = 30°C 34 T1 = 20°C,  = 1.2 × 10 /°C  remains constant V V (II)  = (I)  = R R –5 Now, R = R(1 + ) = R + R × 1.2 × 10 × 30 = 1.00036R From (I) and (II) V V V  = R R 1.00036R V = 1.00036 V (1.00036 V  V ) –2 % change = × 100 = 0.00036 × 100 = 3.6 × 10 V      23.6 CHAPTER 24 KINETIC THEORY OF GASES Volume of mole of gas RT 0.082  273 –3 –2 PV = nRT  V = = = 22.38 ≈ 22.4 L = 22.4 × 10 = 2.24 × 10 m P n=  1 10 3 10 3 PV = = = RT 0.082  273 22.4 22400 23 19 No of molecules = 6.023 × 10 × = 2.688 × 10 22400 –5 V = cm , T = 0°C, P = 10 mm of Hg 1.36  980  10 6  PV ƒgh  V –13 = = = 5.874 × 10 RT RT 8.31  273 23 –13 11 No of moluclues = No × n = 6.023 × 10 × 5.874 × 10 = 3.538 × 10 n= n=  1 10 3 10 3 PV = = RT 0.082  273 22.4 10 3   32 –3 g = 1.428 × 10 g = 1.428 mg 22.4 Since mass is same n1 = n2 = n nR  300 nR  600 P1 = , P2 = V0 2V0 mass = 2V0 2V0 P1 nR  300  = = =1:1 P2 V0 nR  600 600 K –3 V = 250 cc = 250 × 10 –3 –3 –3 –6 –3 P = 10 mm = 10 × 10 m = 10 × 13600 × 10 pascal = 136 × 10 pascal T = 27°C = 300 K 136  10 3  250 PV 136  250 =  10  =  10  8.3  300 RT 8.3  300 136  250 No of molecules =  10    10 23 = 81 × 1017 ≈ 0.8 × 1015 8.3  300 P1 = 8.0 × 10 Pa, P2 = × 10 Pa, T1 = 300 K, Since, V1 = V2 = V n= T2 = ? P1V1 PV  10  V  10  V 1 10  300 = 2  =  T2 = = 375° K T1 T2 300 T2  10 T = 300 K, P=? m = g, V = 0.02 m = 0.02 × 10 cc = 0.02 × 10 L, M = g, m PV = nRT  PV = RT  P × 20 =  0.082  300 M 0.082  300 5 P= = 1.23 atm = 1.23 × 10 pa ≈ 1.23 × 10 pa 20 nRT m RT ƒRT P= =  = V M V M –3 ƒ  1.25 × 10 g/cm R  8.31 × 10 ert/deg/mole T  273 K M= 1.25  10 3  8.31  10  273 ƒRT = = 0.002796 × 10 ≈ 28 g/mol P 13.6  980  76 24.1 V0 300 K Kinetic Theory of Gases 10 T at Simla = 15°C = 15 + 273 = 288 K –2 P at Simla = 72 cm = 72 × 10 × 13600 × 9.8 T at Kalka = 35°C = 35 + 273 = 308 K –2 P at Kalka = 76 cm = 76 × 10 × 13600 × 9.8 PV = nRT m m PM  PV = RT  PM = RT  ƒ = M V RT PSimla  M RTKalka ƒSimla  = ƒKalka RTSimla PKalka  M = 72  10 2  13600  9.8  308 2 288  76  10  13600  9.8 ƒKalka = = 0.987 ƒSimla 1.013 11 n1 = n2 = n nRT nRT , P2 = P1 = V 3V P1 nRT V =  =3:1 P2 V nRT = 72  308 = 1.013 76  288 V V 3V V PT PT P2T P1 - 12 r.m.s velocity of hydrogen molecules = ? –3 T = 300 K, R = 8.3, M = g = × 10 Kg 3RT C= M C=  8.3  300  10  = 1932 m/s ≈1930 m/s Let the temp at which the C = × 1932.6 is T × 1932.6 =   T   10 3  (2 × 1932.6) =   T   10  (2  1932.6)2   10 3 = T   T = 1199.98 ≈ 1200 K  13 Vrms = = 3P ƒ P = 10 Pa = atm, ƒ= 1.77  10 4 10  3  10  10 3 = 1301.8 ≈ 1302 m/s 1.77  10  14 Agv K.E = 3/2 KT –19 3/2 KT = 0.04 × 1.6 × 10 –23 –19  (3/2) × 1.38 × 10 × T = 0.04 × 1.6 × 10 T=  0.04  1.6  10 19  1.38  10  23 = 0.0309178 × 10 = 309.178 ≈ 310 K 8RT  8.3  300 = M 3.14  0.032 Dis tan ce 6400000  T= = = 445.25 m/s Speed 445 25 15 Vavg = 28747 83 km = 7.985 ≈ hrs 3600 –3 16 M = × 10 Kg =  8.3  273 8RT = = 1201.35 M 3.14   10  –27 –24 –24 Momentum = M × Vavg = 6.64 × 10 × 1201.35 = 7.97 × 10 ≈ × 10 Kg-m/s Vavg = 24.2 Kinetic Theory of Gases 8RT  8.3  300 = M 3.14  0.032 8RT1 8RT2 Now, = 2  17 Vavg = T1 = T2 8RT M 18 Mean speed of the molecule = Escape velocity = 8RT = M T= VavgN2  2gr 8RT = 2gr M  9.8  6400000  3.14   10 3 2grM = = 11863.9 ≈ 11800 m/s 8R  8RT M 19 Vavg = VavgH2 2gr = 8RT   28 =  8RT 2 28 = 14 = 3.74 20 The left side of the container has a gas, let having molecular wt M1 Right part has Mol wt = M2 Temperature of both left and right chambers are equal as the separating wall is diathermic 3RT = M1 21 Vmean = 3RT 8RT 8RT  = M1 M2 M2 8RT = M  M1 M 3 =  = = 1.1775 ≈ 1.18 M2 M2 8  8.3  273 = 1698.96 3.14   10  Total Dist = 1698.96 m 1698.96 10 = 1.23 × 10 No of Collisions = 1.38  10  22 P = atm = 10 Pascal –3 T = 300 K, M = g = × 10 Kg  8.3  300 8RT = = 1781.004 ≈ 1780 m/s M 3.14   10  (b) When the molecules strike at an angle 45°, (a) Vavg = Force exerted = mV Cos 45° – (–mV Cos 45°) = mV Cos 45° = m V No of molecules striking per unit area = = 10   10 3  1780 =  1780 Force 2mv  Area = = mV Pr essure 2mV  10 31 = 1.19 × 10–3 × 1031 = 1.19 × 1028 ≈ 1.2 × 1028  10 23 PV PV 23 1 = 2 T1 T2 P1  200 KPa = × 10 pa T1 = 20°C = 293 K 102  V1 V2 = V1 + 2% V1 = 100  P2 = ? T2 = 40°C = 313 K P  102  V1  10  V1  10  313 =  P2 = = 209462 Pa = 209.462 KPa 293 100  313 102  293 24.3 Kinetic Theory of Gases –3 24 V1 = × 10 m , P1V1 = n1R1T1 n= P1 = 1.5 × 10 Pa, P1V1 1.5  10  1 10 3 = R1T1 8.3  400 T1 = 400 K n=  5 M =  32 = 1.4457 ≈ 1.446   –3 P2 = × 10 Pa, V2 = × 10 m , P2V2 = n2R2T2  m1 =  n2 = T2 = 300 K P2 V2 10  10 3 = = = 0.040 R T2 8.3  300   m2 = 0.04 × 32 = 1.285 m = m1 – m2 =1.446 – 1.285 = 0.1608 g ≈ 0.16 g 5 25 P1 = 10 + ƒgh = 10 + 1000 × 10 × 3.3 = 1.33 × 10 pa –3 T1 = T2 = T, V1 = (2 × 10 ) P2 = 10 , V2 = r , r=? P1V1 PV = 2 T1 T2  1.33  10  4 10   r    (2  10  )3 3 = T1 T2 –9  1.33 × × 10 × 10 = 10 × r 26 P1 = atm = × 10 pa T1 = 300 K V1 = 0.002 m , P1V1 = n1RT1 n= r= –3 10.64  10 3 = 2.19 × 10 ≈ 2.2 mm P1V1  10  0.002 = = = 0.1606 RT1 8.3  300  P2 = atm = 10 pa V2 = 0.0005 m , P2V2 = n2RT2  n2 = T2 = 300 K P2 V2 10  0.0005 = =  = 0.02 RT2 8.3  300  8.3 10 n = moles leaked out = 0.16 – 0.02 = 0.14 27 m = 0.040 g, T = 100°C, MHe = g 3 m T = ? U = nRt =   RT 2 M m m Given   RT  12 =   RT M M  1.5 × 0.01 × 8.3 × 373 + 12 = 1.5 × 0.01 × 8.3 × T 58.4385 = 469.3855 K = 196.3°C ≈ 196°C  T = 0.1245 28 PV = constant 2  P1V1 = P2V2 nRT1 nRT2   V12 =  V2 V1 V2  T1 V1 = T2 V2 = TV = T1 × 2V  T2 = T 24.4 Semiconductor devices 32  = 50, Ib = 50 A, V0 =   RG = 50  2/0.5 = 200 V V0 200  = 8000 V a) VG = V0/V1 =  Vi Ib  Ri 50  106   102 -6 b) Vi = Ib  Ri = 50  10   10 = 0.00025 V = 25 mV R 2 = 10  c) Power gain =   RG = 2   2500  Ri 0.5 33 X = ABC  BCA  CAB a) A = 1, B = 0, C = X = b) A = B = C = X = 34 For ABC  BCA B C A 35 LHS = AB  AB = X  X [X = AB] If X = 0, X = If X = 0, X =  + or + =  RHS = (Proved)   ` C P R1=0.5K  2K N N e THE NUCLEUS CHAPTER - 46 M = Amp, f = M/V, mp = 1.007276 u 1/3 –15 1/3 –27 R = R0A = 1.1  10 A , u = 1.6605402  10 kg = A  1.007276  1.6605402  10 27 /  3.14  R3 14 ‘f’ in CGS = Specific gravity =  10 = 0.300159  10 18 =  10 17 kg/m M M  1030 1 V    1013   1014 17 v f 0.6 2.4  10 V = 4/3 R 1 3   1014 = 4/3   R  R =    1014 6  100  R =   1012  4  R = ½  10  3.17 = 1.585  10 m = 15 km Let the mass of ‘’ particle be xu ‘’ particle contains protons and neutrons  Binding energy = (2  1.007825 u   1.00866 u – xu)C = 28.2 MeV (given)  x = 4.0016 u. 7 Li + p  l +  + E ; Li = 7.016u  = He = 4.0026u ; p = 1.007276 u E = Li + P – 2 = (7.016 + 1.007276)u – (2  4.0026)u = 0.018076 u  0.018076  931 = 16.828 = 16.83 MeV. B = (Zmp + Nmn – M)C Z = 79 ; N = 118 ; mp = 1.007276u ; M = 196.96 u ; mn = 1.008665u B = [(79  1.007276 + 118  1.008665)u – Mu]c = 198.597274  931 – 196.96  931 = 1524.302094 so, Binding Energy per nucleon = 1524.3 / 197 = 7.737 238 234 a) U 2He + Th E = [Mu – (NHC + MTh)]u = 238.0508 – (234.04363 + 4.00260)]u = 4.25487 Mev = 4.255 Mev 238 234 b) E = U – [Th + 2n0 + 2p1] = {238.0508 – [234.64363 + 2(1.008665) + 2(1.007276)]}u = 0.024712u = 23.0068 = 23.007 MeV 223 209 14 Ra = 223.018 u ; Pb = 208.981 u ; C = 14.003 u 223 209 14 Ra  Pb + C 223 209 14 m = mass Ra – mass ( Pb + C)  = 223.018 – (208.981 + 14.003) = 0.034 Energy = M  u = 0.034  931 = 31.65 Me. EZ.N  EZ–1, N + P1  EZ.N  EZ–1, N + 1H [As hydrogen has no neutrons but protons only] E = (MZ–1, N + NH – MZ,N)c  f= E2N = EZ,N–1 + 10 n 2 Energy released = (Initial Mass of nucleus – Final mass of nucleus)c = (MZ.N–1 + M0 – MZN)c 10 P32  S32 + 0v  10 Energy of antineutrino and -particle = (31.974 – 31.972)u = 0.002 u = 0.002  931 = 1.862 MeV = 1.86 – 11 In  P + e We know : Half life = 0.6931 /  (Where  = decay constant) –4 Or  = 0.6931 / 1460 = 8.25  10 S [As half life = 14 = 14  60 sec] 2 Energy = [Mn – (MP + Me)]u = [(Mnu – Mpu) – Mpu]c = [0.00189u – 511 KeV/c ] 2 = [1293159 ev/c – 511000 ev/c ]c = 782159 eV = 782 Kev. 46.1 The Nucleus 12 13 226 58 Ra  24   222 26 Rn 19 O 0  19 F  n  v 13 25 Al 25  12 MG  01e  00 v 64 Cu  64Ni  e  v Emission of nutrino is along with a positron emission a) Energy of positron = 0.650 MeV Energy of Nutrino = 0.650 – KE of given position = 0.650 – 0.150 = 0.5 MeV = 500 Kev b) Momentum of Nutrino = 14 a) 500  1.6  10 19  108 19 K 40  20 Ca40  1e0  v 19 K 40  18 Ar 40  1e0  v 19 K 40  1e0  18 Ar 40 19 K 40  20 Ca40  1e0  v –22  103 J = 2.67  10 kg m/s b) Q = [Mass of reactants – Mass of products]c = [39.964u – 39.9626u] = [39.964u – 39.9626]uc = (39.964 – 39.9626) 931 Mev = 1.3034 Mev 19 K 40  18 Ar 40  1e0  v Q = (39.9640 – 39.9624)uc = 1.4890 = 1.49 Mev 19 K 40  1e  18 Ar 40 Qvalue = (39.964 – 39.9624)uc 15 Li  n Li  73Li ; 73 Li  r  83Li  84Be  e  v  4Be  42He  24He + 16 C  B +  + v mass of C = 11.014u ; mass of B = 11.0093u Energy liberated = (11.014 – 11.0093)u = 29.5127 Mev For maximum K.E of the positron energy of v may be assumed as  Maximum K.E of the positron is 29.5127 Mev 17 Mass 238Th = 228.028726 u ; 224Ra = 224.020196 u ;  = 24He  4.00260u 238 224 Th  Ra* +  224 Ra*  Ra + v(217 Kev) 224 Now, Mass of Ra* = 224.020196  931 + 0.217 Mev = 208563.0195 Mev 226Th 224 – E( Ra* + ) KE of  = E = 228.028726  931 – [208563.0195 + 4.00260  931] = 5.30383 Mev= 5.304 Mev 12 12 + 18 N  C* + e + v 12 12 C*  C + v(4.43 Mev) 12 12 + Net reaction : N  C + e + v + v(4.43 Mev) + 12 12 Energy of (e + v) = N – (c + v) = 12.018613u – (12)u – 4.43 = 0.018613 u – 4.43 = 17.328 – 4.43 = 12.89 Mev Maximum energy of electron (assuming energy for v) = 12.89 Mev 19 a) t1/2 = 0.693 /  [  Decay constant]  t1/2 = 3820 sec = 64 b) Average life = t1/2 / 0.693 = 92 –t c) 0.75 = e  In 0.75 = – t  t = In 0.75 / –0.00018 = 1598.23 sec. 20 a) 198 grams of Ag contains  N0 atoms 224 g of Ag contains  N0/198  g =  1023  1 10 6 atoms 198 46.2 The Nucleus Activity = N = 0.963 0.693   1017 N = disintegrations/day t1/ 198  2.7 0.693   1017 0.693   1017 disintegration/sec = curie = 0.244 Curie 198  2.7  3600  24 198  2.7  36  24  3.7  1010 A0 0.244  = 0.0405 = 0.040 Curie. b) A = 2t1/ 2 2.7 21 t1/2 = 8.0 days ; A0 = 20  Cl a) t = 4.0 days ;  = 0.693/8 = = 20  10  e( 0.693 / 8)4 = 1.41  10 Ci = 14  Ci 0.693 –6 b)  = = 1.0026  10   24  3600 –18 –1 22  = 4.9  10 s 1 238 a) Avg life of U =    10 18 sec  4.9  1018 4.9 = 6.47  10 years 0.693 0.693  = 4.5  10 years b) Half life of uranium =  4.9  1018 A A c) A = t / t0   2t / t1/ = 22 = A 1/ 23 A = 200, A0 = 500, t = 50 –t –50 60 A = A0 e or 200 = 500  e    –4   = 3.05  10 s 0.693 0.693 = 2272.13 sec = 38 min. b) t1/2 =   0.000305 24 A0 =  10 disintegration / sec A =  10 dis/sec ; t = 20 hours A A A = t / t0  2t / t1/   2t / t1/  A' 1/ 1/2  t / t1/ =  t = t/2 = 20 hours / = 10 hours  –t A = A0e A0 t / t1/ –6 –5  106 = 0.00390625  10 = 3.9  10 dintegrations/sec 2100 / 10 25 t1/2 = 1602 Y ; Ra = 226 g/mole ; Cl = 35.5 g/mole mole RaCl2 = 226 + 71 = 297 g 297g = mole of Ra A =  A   0.1 6.023  1023 22 = 0.02027  10  0.1 mole of Ra = 297 297 –11  = 0.693 / t1/2 = 1.371  10 –11 20 9 Activity = N = 1.371  10  2.027  10 = 2.779  10 = 2.8  10 disintegrations/second. 26 t1/2 = 10 hours, A0 = ci 0.1 g = 0.693 –t Activity after hours = A0 e = 1 e 10 th No of atoms left after hour, A9 = N9  N9 = 9 = 0.5359 = 0.536 Ci A 0.536  10  3.7  1010  3600 10 13 = 28.6176  10  3600 = 103.023  10   0.693 –t Activity after 10 hours = A0 e = 1 e th No of atoms left after 10 hour A10 = N10 0.693 9 10 = 0.5 Ci 46.3 The Nucleus A10 0.5  3.7  1010  3600 10 13  = 26.37  10  3600 = 96.103  10  0.693 /10 13 13 No.of disintegrations = (103.023 – 96.103)  10 = 6.92  10 27 t1/2 = 14.3 days ; t = 30 days = month As, the selling rate is decided by the activity, hence A0 = 800 disintegration/sec –t [ = 0.693/14.3] We know, A = A0e A = 800  0.233669 = 186.935 = 187 rupees. 28 According to the question, the emission rate of  rays will drop to half when the + decays to half of its original amount And for this the sample would take 270 days  The required time is 270 days + + 29 a) P  n + e + v Hence it is a  decay b) Let the total no of atoms be 100 N0 Carbon Boron 10 N0 Initially 90 N0 Finally 10 N0 90 N0  N10 = 0.693 –t t Now, 10 N0 = 90 N0 e  1/9 = e 20.3 [because t1/2 = 20.3 min] 0.693 2.1972  20.3  In  = 64.36 = 64 min. tt 20.3 0.693 23 30 N =  10 ; t1/2 = 12.3 years dN 0.693 0.693 a) Activity =  n  N   1023 dis/year dt t1/ 12.3 14 = 7.146  10 dis/sec dN 14 b)  7.146  10 dt 14 17 19 No.of decays in next 10 hours = 7.146  10  10  36 = 257.256  10 = 2.57  10 0.693 –t 23 6.16 23 = 2.82  10 = No.of atoms remained c) N = N0 e =  10  e 20.3 23 23  No of atoms disintegrated = (4 – 2.82)  10 = 1.18  10 31 Counts received per cm = 50000 Counts/sec 16 N = N3o of active nucleic =  10 Total counts radiated from the source = Total surface area  50000 counts/cm 4 =  3.14   10   10 = 6.28  10 Counts = dN/dt cm2 dN We know,  N dt 6.28  109 1m –7 –7 –1 = 1.0467  10 = 1.05  10 s   1016 32 Half life period can be a single for all the process It is the time taken for 1/2 of the uranium to convert to lead Or  = No of atoms of U 238 =  1023   103 12 20 =  10 20 = 0.05042  10 238 238  1023  0.6  10 3 3.6   1020 206 206 3.6   12 20  Initially total no of uranium atoms =    10 = 0.06789  235 206  No of atoms in Pb = 0.693 0.693 –t N = N0 e  N = N0 e t / t1/  0.05042 = 0.06789 e 4.4710 0.693t  0.05042   log    0.06789  4.47  109  t = 1.92  10 years. 46.4 The Nucleus 33 A0 = 15.3 ; A = 12.3 ; t1/2 = 5730 year = 0.6931 0.6931 1  yr T1/ 5730 Let the time passed be t, We know A = A et  0.6931  t  12.3 = 15.3  e 5730  t = 1804.3 years. 34 The activity when the bottle was manufactured = A0 Activity after years = A e 0.693 8 12.5 Let the time of the mountaineering = t years from the present A = A0e 0.693 t 12.5 ; A = Activity of the bottle found on the mountain A = (Activity of the bottle manufactured years before)  1.5%  A0e 0.693 12.5 = A0e 0.693 8 12.5  0.015 0.693 0.6938 t  In[0.015] 12.5 12.5  0.05544 t = 0.44352 + 4.1997  t = 83.75 years –1 35 a) Here we should take R0 at time is t0 = 30  10 s   30  109  i) In(R0/R1) = In   30  109  =   30 25  30  109 ii) In(R0/R2) = In   16  10   = 0.63  Count rate R(109 s–1) 20  30  109 iii) In(R0/R3) = In    10   = 1.35  10  30  10 iv) In(R0/R4) = In   3.8  10   = 2.06  15 25  30  109  v) In(R0/R5) = In    109  = 2.7   –1 b)  The decay constant  = 0.028 c)  The half life period = t1/2 0.693 0.693 = 25 min.   0.028 9 36 Given : Half life period t1/2 = 1.30  10 year , A = 160 count/s = 1.30  10  365  86400 0.693  A = N  160 = N t1/  t1/2 =  N= 160  1.30  365  86400  109 18 = 9.5  10 0.693 23  6.023  10 No of present in 40 grams 6.023  1023 = 40 g  = 18 40 6.023  1023 40  9.5  1018 –4 = 6.309  10 = 0.00063 6.023  1023  The relative abundance at 40 k in natural potassium = (2  0.00063  100)% = 0.12%  9.5  10 present in = 50 75 100 Time t (Minute) 46.5 The Nucleus -1/2 37 a) P + e  n + v neutrino [a  4.95  10 s ; b  1] b) f = a(z – b)  c /  = 4.95  10 (79 – 1) = 4.95  10  78  C/ = (4.95  78)  10  =  108 14903.2  1014 =  10 –5 –6 –4  10 =  10 14 m = 20 pm. dN dN R R= dt dt Given after time t >> t1/2, the number of active nuclei will become constant i.e (dN/dt)present = R = (dN/dt)decay  R = (dN/dt)decay  R = N [where,  = Radioactive decay constant, N = constant number] Rt1/ 0.693  R= (N)  Rt1/2 = 0.693 N  N =  t1/ 0.693 38 Given : Half life period = t1/2, Rate of radio active decay = 39 Let N0 = No of radioactive particle present at time t = N = No of radio active particle present at time t –t  N = N0 e [ - Radioactive decay constant] –t –t  The no.of particles decay = N0 – N = N0 – N0e = N0 (1 – e ) We know, A0 = N0 ; R = N0 ; N0 = R/ From the above equation R –t (substituting the value of N0) N = N0 (1 – e ) = (1  e t )  23 40 n = mole =  10 atoms, t1/2 = 14.3 days t = 70 hours, dN/dt in root after time t = N 0.69370 –t 23 23 23 =  10  e 14.324 =  10  0.868 = 5.209  10 23 5.209  1023  0.693  0.010510 dis/hour 14.324 3600 –6 23 17 = 2.9  10  10 dis/sec = 2.9  10 dis/sec  1ci  Fraction of activity transmitted =    100%  2.9  1017  N = No e  1 3.7  108  –11    2.9  1011  100  % = 1.275  10 %   41 V = 125 cm3 = 0.125 L, P = 500 K pa = atm T = 300 K, t1/2 = 12.3 years = 3.82  10 sec Activity =   N  0.125 23 22 N = n  6.023  10 =  6.023  1023 = 1.5  10 atoms 8.2  10 2   102 0.693 –8 –9 –1 = = 0.1814  10 = 1.81  10 s 3.82  108 –9 22 Activity = N = 1.81  10  1.5  10 = 2.7  10 disintegration/sec = 42 2.7  1013 3.7  1010 Ci = 729 Ci. 212 83 Bi 208  81 Ti  24He( ) 212 83 Bi 212 212  84 Bi  84 P0  e  t1/2 = h Time elapsed = hour 212 Present = g at t = Bi 212  at t = Bi Present = 0.5 g Probability -decay and -decay are in ratio 7/13  Tl remained = 0.175 g  P0 remained = 0.325 g 46.6 The Nucleus 108 110 43 Activities of sample containing Ag and Ag isotopes = 8.0  10 disintegration/sec a) Here we take A =  10 dis./sec  i) In (A1/ A 01 ) = In (11.794/8) = 0.389 12 ii) In (A2/ A 02 ) = In(9.1680/8) = 0.1362 10 iii) In (A3/ A 03 ) = In(7.4492/8) = –0.072 iv) In (A4/ A 04 ) = In(6.2684/8) = –0.244 v) In(5.4115/8) = –0.391 vi) In(3.0828/8) = –0.954 vii) In(1.8899/8) = –1.443 viii) In(1.167/8) = –1.93 ix) In(0.7212/8) = –2.406 b) The half life of 110 Ag from this part of the plot is 24.4 s 110 c) Half life of Ag = 24.4 s  decay constant  = 0.693/24.4 = 0.0284  t = 50 sec, –t –0.028450 The activity A = A0e =  10  e = 1.93  10 d) 20 40 60 80 100 200 300 400 500 Time O 20 40 60 80 108 e) The half life period of Ag from the graph is 144 s. 44 t1/2 = 24 h tt 24   t1/2 =  = 4.8 h t1  t 24  A0 = rci ; A = rci A rci t  A = t / t0  rci = t / 4.8h  =  t = 4.8 h 24.8h 2 1/ 45 Q = qe  t / CR ; A = A0e –t Energy 1q2  e 2t / cR  Activity CA e t Since the term is independent of time, so their coefficients can be equated, 2t 2  = t or,  = or,  or, R = (Proved) So,  CR CR CR C 46 R = 100  ; L = 100 mH After time t, i = i0 (1  e t / Lr ) i i0 (1  e  tR / L )  N N0 et –t N = N0 (e ) i/N is constant i.e independent of time Coefficients of t are equal –R/L = –  R/L = 0.693/t1/2 –3 –4 = t1/2 = 0.693  10 = 6.93  10 sec. 235 23 47 g of ‘I’ contain 0.007 g U So, 235 g contains 6.023  10 atoms So, 0.7 g contains 6.023  1023  0.007 atom 235 atom given 200 Mev So, 0.7 g contains 6.023  10 23  0.007  200  106  1.6  10 19 –8 J = 5.74 10 J 235 48 Let n atoms disintegrate per second –19 Total energy emitted/sec = (n  200  10  1.6  10 ) J = Power 300 MW = 300  10 Watt = Power 46.7 The Nucleus 6 300  10 = n  200  10  1.6  10 3  n=  1019 =  1019  1.6 3.2  10 23 –19 atoms are present in 238 grams 238   1019 –4 = 3.7  10 g = 3.7 mg  1019 atoms are present in 3.2  1023  3.2 49 a) Energy radiated per fission =  10 ev –12 Usable energy =  10  25/100 =  10 ev =  1.6  10 8 Total energy needed = 300  10 =  10 J/s  108 20 = 0.375  10  1.6  1012 20 24 No of fission per day = 0.375  10  3600  24 = 3.24  10 fissions 24 b) From ‘a’ No of atoms disintegrated per day = 3.24  10 23 We have, 6.023  10 atoms for 235 g 235 for 3.24  1024 atom =  3.24  1024 g = 1264 g/day = 1.264 kg/day 23 6.023  10 No of fission per second = 50 a) 2 H  1H  13H  11H Q value = 2M(12 H) = [M(13 H)  M(13 H)] = [2  2.014102 – (3.016049 + 1.007825)]u = 4.0275 Mev = 4.05 Mev b) 2 H  1H  32H  n Q value = 2[M(12 H)  M(32 He)  Mn ] = [2  2.014102 – (3.016049 + 1.008665)]u = 3.26 Mev = 3.25 Mev c) H  1H  24H  n Q value = [M(12 H)  M(13 He)  M( 24 He)  Mn ] = (2.014102 + 3.016049) – (4.002603 + 1.008665)]u = 17.58 Mev = 17.57 Mev Kq1q2  109  (2  1.6  1019 )2 = r r –23 1.5 KT = 1.5  1.38  10  T 51 PE = Equating (1) and (2) 1.5  1.38  10 –23 …(1) …(2) T=  109  10.24  10 38  10 15  109  10.24  10 38 10 = 22.26087  10 K = 2.23  10 K  10 15  1.5  1.38  10 23 4  Be 52 H + H  4.0026 u M( H)  8.0053 u M( Be) Q value = [2 M( H) – M( Be)] = (2  4.0026 – 8.0053) u = –0.0001 u = –0.0931 Mev = –93.1 Kev 23 53 In 18 g of N0 of molecule = 6.023  10  T= 6.023  1026 25 = 3.346  10 18 26  % of Deuterium = 3.346  10  99.985 25 Energy of Deuterium = 30.4486  10 = (4.028204 – 3.016044)  93 = 942.32 ev = 1507  10 J = 1507 mJ In 100 g of N0 of molecule =  46.8 THE SPECIAL THEORY OF RELATIVITY CHAPTER - 47 S = 1000 km = 10 m The process requires minimum possible time if the velocity is maximum We know that maximum velocity can be that of light i.e =  10 m/s Distance 106   s Speed  108 300 ℓ = 50 cm, b = 25 cm, h = 10 cm, v = 0.6 c a) The observer in the train notices the same value of ℓ, b, h because relativity are in due to difference in frames b) In different frames, the component of length parallel to the velocity undergoes contraction but the perpendicular components remain the same So length which is parallel to the x-axis changes and breadth and height remain the same So, time = e = e  V2 C  50  C2 = 50  0.36 = 50  0.8 = 40 cm The lengths observed are 40 cm  25 cm  10 cm L=1m a) v  10 m/s L = 1   1010  1016 b) v = x 10 m/s L = 1   1012  1016 c) v =  10 m/s   10 6 = 0.9999995 m   10 4 = 0.99995 m  1014   10 2 = 0.9949 = 0.995 m  1016 v = 0.6 cm/sec ; t = sec a) length observed by the observer = vt  0.6   10  1.8  10 m L = 1  (0.6)2 C2 b) ℓ =   v / c  1.8  10 =   (0.6)2 C2 C2 1.8  108 = 2.25  10 m/s 0.8 The rectangular field appears to be a square when the length becomes equal to the breadth i.e 50 m i.e L = 50 ; L = 100 ; v = ? C =  10 m/s ℓ0 = We know, L = L  v / c  50 = 100  v / c  v = / 2C = 0.866 C L0 = 1000 km = 10 m v = 360 km/h = (360  5) / 18 = 100 m/sec 10  100  a) h = h0  v / c  10   = 10   10   10   10  Solving change in length = 56 nm b) t = L/v = 56 nm / 100 m = 0.56 ns. 47.1 The Special Theory of Relativity v = 180 km/hr = 50 m/s t = 10 hours let the rest dist be L A B L = L  v / c  L = 10  180 = 1800 k.m 1802 1800 = L  (3  105 )2 –14 or, 1800  1800 = L(1 – 36  10 3.24  10 or, L = ) = 1800 + 25  10 14 –12  36  10 or 25 nm more than 1800 km b) Time taken in road frame by Car to cover the dist = 1.8  106  25  10 9 50 –8 = 0.36  10 +  10 = 10 hours + 0.5 ns a) u = 5c/13 t t = 1 v / c 1y  1  25c 169c y  13 13  y 12 12 The time interval between the consecutive birthday celebration is 13/12 y b) The fried on the earth also calculates the same speed. The birth timings recorded by the station clocks is proper time interval because it is the ground frame That of the train is improper as it records the time at two different places The proper time interval T is less than improper i.e T = v T Hence for – (a) up train  Delhi baby is elder (b) down train  Howrah baby is elder. 10 The clocks of a moving frame are out of synchronization The clock at the rear end leads the one at from by L0 V/C where L0 is the rest separation between the clocks, and v is speed of the moving frame Thus, the baby adjacent to the guard cell is elder 11 v = 0.9999 C ; t = One day in earth ; t = One day in heaven v= 1 v / c  1 (0.9999) C C2  = 70.712 0.014141782 t = v t ; Hence, t = 70.7 days in heaven. 12 t = 100 years ; V = 60/100 K ; C = 0.6 C t = t 1 V / C 100y  1 (0.6) C C2  100y = 125 y. 0.8 13 We know f = f  V / C2 f = apparent frequency ; f = frequency in rest frame v = 0.8 C f = 1 0.64C2 C  0.36 = 0.6 s –1 The Special Theory of Relativity 14 V = 100 km/h, t = Proper time interval = 10 hours t 10  3600  t = 2 1 V / C  1000  1    36   10    t – t = 10  3600   1 1000   1     36   108   By solving we get, t – t = 0.154 ns  Time will lag by 0.154 ns. 15 Let the volume (initial) be V V = V/2 So, V/2 = v  V / C2  C/2 = C2  V  C2/4 = C2 – V2 C2 3  C V= C 4 16 d = cm, v = 0.995 C  V = C2  a) time in Laboratory frame = 1 10 2 d 1 10 2  v 0.995C = 33.5  10 0.995   108 b) In the frame of the particle = t = t  V / C2  –12 = 33.5 PS 33.5  10 12  (0.995)2 = 335.41 PS –2 17 x = cm =  10 m ; K = 500 N/m, m = 200 g –4 Energy stored = ½ Kx = ½  500  10 = 0.025 J Increase in mass = 0.025 C  Fractional Change of max = 0.025  1016 0.025   10  10 1 18 Q = MS    4200 (100 – 0) = 420000 J E = (m)C E Q 420000  m =   C C (3  108 )2 –12 16 = 0.01388  10 –12 = 4.66  10 = 4.7  10 kg 19 Energy possessed by a monoatomic gas = 3/2 nRdt Now dT = 10, n = mole, R = 8.3 J/mol-K E = 3/2  t  8.3  10 1.5  8.3  10 124.5  C2  1015 –16 –15 = 1383  10 = 1.38  10 Kg 20 Let initial mass be m ½ mv = E Loss in mass =  12   m  50 m    18  m = E/C  E= –16 –8 = 1.4  10 The Special Theory of Relativity m  50 m 50  m   10 81 1016 –16 –17  0.617  10 = 6.17  10 21 Given : Bulb is 100 Watt = 100 J/s So, 100 J in expended per sec Hence total energy expended in year = 100  3600  24  365 = 3153600000 J Total energy 315360000 Change in mass recorded =  C2  1016 –16 –8 = 3.504  10  10 kg = 3.5  10 Kg  m =  16 22 I = 1400 w/m Power = 1400 w/m  A 11 = (1400  4R )w = 1400  4  (1.5  10 ) 22 = 1400  4  (1/5)  10 a) sun E mC2 m E / t    t t t C 1400    2.25  1022 66 9 = 1696  10 = 4.396  10 = 4.4  10  1016 b) 4.4  109 Kg disintegrates in sec C =  10 30 Kg disintegrates in  1030 4.4  109 sec   1 1021 –8 21 13 =   = 1.44  10  10 y = 1.44  10 y  2.2  365  24  3600  –31 23 Mass of Electron = Mass of positron = 9.1  10 Kg Both are oppositely charged and they annihilate each other Hence, m = m + m =  9.1  10–31 Kg Energy of the resulting  particle = m C =  9.1  10 –31   10 16 J=  9.1  10 15 1.6  10 19 = 102.37  10 ev = 1.02  10 ev = 1.02 Mev. –31 24 me = 9.1  10 , v0 = 0.8 C a) m = Me  V / C2  9.1 1031  0.64C2 / C2 –31  ev 9.1 10 31 0.6 –31 = 15.16  10 Kg = 15.2  10 Kg 2 b) K.E of the electron : mC – meC = (m – me) C –31 –31 –31 18 = (15.2  10 – 9.1  10 )(3  10 ) = (15.2  9.1)   10  10 J –15 –14 –14 = 54.6  10 J = 5.46  10 J = 5.5  10 J c) Momentum of the given electron = Apparent mass  given velocity –31 –23 = 15.2  10 – 0.8   10 m/s = 36.48  10 kg m/s –22 = 3.65  10 kg m/s 25 a) ev – m0C = m0 C2 1 = V C2 9.1  1031  1016 1 0.36C C2  ev – 9.1  10 –31   10  eV – 9.1   10 –15 16 R The Special Theory of Relativity 9.1  10 15 9.1  10 15 –15  eV – 9.1   10 =  0.8 1.6 =  9.1   81.9   9.1   1015 = eV   81.9   10 15  eV =   1.6   1.6  –15 eV = 133.0875  10  V = 83.179  10 = 831 KV m0 C2 b) eV – m0C = 1  eV – 81.9  10 –15  eV = 12.237  10  V=  eV – 9.1   10 V2 C2 m0 C2 1 V C2 +m0C =  eV = 372.18  10 9.1  10 15 = 76.48 kV 2 1 = 1 V = 0.99 C = eV – m0C = 16 –15 1.6  10 19 m0 C2   10 9.1  10 15  0.435 = 12.237  10 15  eV = –19 –15 V2 C2 9.1 10 31   1016 2  (0.99) V= 372.18  20 15 1.6  1019  9.1 1031   1016 = 272.6  10  V = 2.726  10 = 2.7 MeV 26 a) m0 C2 V2 1 C – m0C = 1.6  10 –19   –19  m0 C2   1 = 1.6  10 2  1 V / C   1  V / C2 1 = 1.6  10 19 9.1 10 31   1016  V = C  0.001937231 =  0.001967231  = 5.92  10 m/s b) m0 C2 1 V C2 – m0C = 1.6  10 –19  10  10   –15  m0 C2   1 = 1.6  10 2  1 V / C   1  V / C2 1 = 1.6  10 15 9.1  1015  V = 0.584475285  10 = 5.85  10 m/s –19 –12 c) K.E = 10 Mev = 10  10 eV = 10  1.6  10 J = 1.6  10 J  m0 C2 V 1 C 2 – m0C = 1.6  10 16  V = 999991359  10 –12 J  V = 2.999987038  10 0.81C2 C2 The Special Theory of Relativity 27 m = m – m0 = 2m0 – m0 = m0 –31 16 Energy E = m0c = 9.1  10   10 J E in e.v = 9.1  10 15 1.6  10 19 = 51.18  10 ev = 511 Kev  m0 C2   m0 C2   mv    V  1  C  28  = 0.01 m0 v   v2 V2 V6 )  m0C2        m0 C (1  2 4 2C C C   mv = 0.1     m0 v   V 15 V4 m0 v  m0  m0  m0 v 2 96 C C  = 0.01 m0 v V4 15 V = 0.01  4C 96  C4 Neglecting the v term as it is very small   V2 V2 = 0.01  = 0.04 / C2 C2 0.2   108 1.732 = 0.346  10 m/s = 3.46  10 m/s  V/C = 0.2 / = V = 0.2 / C =  ... Cp1 =  1  1 In Gas (2) we have, , Cp2 (Sp Heat at const ‘P’), Cv2 (Sp Heat at const ‘V’), n2 (No of moles) Cp R R  & Cp2 – Cv2 = R  Cv2 – Cv2 = R  Cv2 ( – 1) = R  Cv2 = & Cp2 = Cv... CB3  CA = CB And when B & C are mixed   = 420 0 + CB When A & c are mixed, if T is the common temperature of mixture MCA (T – 12) = MCC (28 – T) MCB (23 – 19)= MCC (28 – 23 )  4CB = 5CC  CC =... 19 C The temp of C = 28 C The temp of  A + B = 16° The temp of  B + C = 23 ° In accordance with the principle of caloriemetry when A & B are mixed …(1) MCA (16 – 12) = MCB (19 – 16)  CA4 = CB3