physics_nuclei
 Nuclei The atomic mass unit (u), defined as 1/12th of the mass of the carbon ( 12C ) $\left({\phantom{\rule{0.22em}{0ex}}}^{12}C\phantom{\rule{0.22em}{0ex}}\right)\phantom{\rule{0.22em}{0ex}}$atom.1 u =mass of one 12C atom12 $1\phantom{\rule{0.22em}{0ex}}u\phantom{\rule{0.22em}{0ex}}=\frac{mass\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}one{\phantom{\rule{0.22em}{0ex}}}^{12}C\phantom{\rule{0.22em}{0ex}}atom}{12}\phantom{\rule{0.44em}{0ex}}$1 u=1.660539×10-27 kg $1\phantom{\rule{0.22em}{0ex}}u=1.660539×{10}^{-27}\phantom{\rule{0.22em}{0ex}}kg\phantom{\rule{0.22em}{0ex}}$ A free neutron, is unstable. It decays into a proton, an electron and a anti neutrino (another elementary particle), and has a mean life of about 1000s. Neutron is stable inside the nucleus. Isobar: Nuclides with same mass number and different atomic number is called isobar. Isotone: Nuclides with same neutron number and different atomic number is called isotone. Radius of Nucleus of mass number AR=R0A13$R={R}_{0}{A}^{\frac{1}{3}}$R0=1.2×10-15 m ${R}_{0}=1.2×{10}^{-15}\phantom{\rule{0.22em}{0ex}}m\phantom{\rule{0.22em}{0ex}}$ Volume of the nucleus is proportional to "Mass Number" Density of nucleus is constant, it is independent of "mass number" Types of radioactive decay: 𝛼-decay :$\phantom{\rule{0.22em}{0ex}}𝛼-decay\phantom{\rule{0.22em}{0ex}}:$ In this Helium nucleus is emitted 𝛽-decay :$\phantom{\rule{0.22em}{0ex}}𝛽-decay\phantom{\rule{0.22em}{0ex}}:$ In this electron or positrons are emitted 𝛾-decay :$\phantom{\rule{0.22em}{0ex}}𝛾-decay\phantom{\rule{0.22em}{0ex}}:$ In this high energy photons are emitted Alpha-Decay:When a nucleus undergoes alpha decay it transforms to a different nucleus by emitting alpha particles.AZX→A-4Z-2X+42He ${Z}_{A}^{}X\to {Z-2}_{A-4}^{}X+{2}_{4}^{}He\phantom{\rule{0.22em}{0ex}}$ Beta Decay:A nucleus that decays spontaneously by emitting an electron or a positron is said to undergo beta decay. Antineutrino and neutrino are neutral particlesAntineutrino and neutrino have little or no mass ⏨ v =antineutrino$\overline{\phantom{\rule{0.22em}{0ex}}v\phantom{\rule{0.22em}{0ex}}}=antineutrino$ v=neutrino$v=neutrino$ Beta Minus DecayA neutron transforms into a proton within the nucleus n→p+e-+⏨ v $n\to p+{e}^{-}+\overline{\phantom{\rule{0.22em}{0ex}}v\phantom{\rule{0.22em}{0ex}}}\phantom{\rule{0.44em}{0ex}}$ Beta Plus DecayA proton transforms into a neutron within the nucleus p→n+e++v $p\to n+{e}^{+}+v\phantom{\rule{0.44em}{0ex}}$ Gamma decayWhen a nucleus is in an excited state, it can make a transition to a lower energy state by the emission of electromagnetic radiation. As the energy differences between levels in a nucleus are of the order of MeV, the photons emitted by the nuclei have MeV energies and are called gamma rays. Most radionuclides after an alpha decay or a beta decay leave the daughter nucleus in an excited state.The daughter nucleus reaches the ground state by a single transition or sometimes by successivetransitions by emitting one or more gamma rays. Law of radioactive decay:Number of nuclei undergoing the decay per unit time is proportional to the total number of nuclei in the sample 𝛥N𝛥t∝N$\phantom{\rule{0.22em}{0ex}}\frac{𝛥N}{𝛥t}\propto N$dNdt=-𝜆N$\frac{dN}{dt}=-𝜆N$ N(t)=N0 e-𝜆t $N\left(t\right)={N}_{0}\phantom{\rule{0.22em}{0ex}}{e}^{-𝜆t}\phantom{\rule{0.44em}{0ex}}$ where, 𝜆$𝜆$ is called the radioactive decay constant or disintegration constant. Decay rate at instant t: R=R0 e-𝜆t $R={R}_{0}\phantom{\rule{0.22em}{0ex}}{e}^{-𝜆t}\phantom{\rule{0.44em}{0ex}}$R=𝜆N $R=𝜆N\phantom{\rule{0.22em}{0ex}}$where, N=number of radioactive nuclie that have not undergone decay𝜆=Disintegration $𝜆=Disintegration\phantom{\rule{0.22em}{0ex}}$ constantRo${R}_{o}$=radioactive decay rate at t=0 The total decay rate R of a sample of one or more radionuclides is called the activity of that sample. SI unit for activity : becquerel.1 becquerel = 1Bq = 1 decay per second 1 curie=1 Ci=3.7×1010 Bq $1\phantom{\rule{0.22em}{0ex}}curie=1\phantom{\rule{0.22em}{0ex}}Ci=3.7×{10}^{10}\phantom{\rule{0.22em}{0ex}}Bq\phantom{\rule{0.22em}{0ex}}$ Half Life of nuclideT1/2=ln 2𝜆=0.693𝜆 ${T}_{1/2}=\frac{ln\phantom{\rule{0.22em}{0ex}}2}{𝜆}=\frac{0.693}{𝜆}\phantom{\rule{0.44em}{0ex}}$ Disintegration constant=𝜆$Disintegration\phantom{\rule{0.22em}{0ex}}constant=𝜆$ Alpha Decay: Transforms into different nucleus by emitting alpha particles. 23892U→ 23490Th+ 42He $\phantom{\rule{0.44em}{0ex}}{92}_{238}^{}U\to \phantom{\rule{0.44em}{0ex}}{90}_{234}^{}Th+\phantom{\rule{0.44em}{0ex}}{2}_{4}^{}He\phantom{\rule{0.44em}{0ex}}$ AZX→ A-2Z-2Y+ 42He $\phantom{\rule{0.44em}{0ex}}{Z}_{A}^{}X\to \phantom{\rule{0.44em}{0ex}}{Z-2}_{A-2}^{}Y+\phantom{\rule{0.44em}{0ex}}{2}_{4}^{}He\phantom{\rule{0.44em}{0ex}}$ Q value or disintegration energy=Q=(mX-mY-mHe)c2$=Q=\left({m}_{X}-{m}_{Y}-{m}_{He}\right){c}^{2}$ If neutrons and protons are brought together to form a nucleus , an energy Eb${E}_{b}$ will be released in this process.This energy Eb${E}_{b}$ is called binding energy. The nucleus could be split into nucleons by supplying external energy equal to the binding energy. Fission: A very heavy nucleus (A=240) has lower binding energy per nucleon compared to that of a nucleus with A = 120.If a nucleus A = 240 breaks into two A = 120 nuclei, nucleons get more tightly bound. This implies energy would be released in the process. This is called fission. FusionIf two very light nuclei (A⩽10$A⩽10$) joining to form a heavier nucleus. The binding energy per nucleon of the fused heavier nuclei is more than the binding energy per nucleon of the lighter nuclei.This means that the final system is more tightly bound than the initial system. Again energy would be released in such a process which is known as fusion. Energy can be released if two light nuclei combine to form a single larger nucleus. This process is called nuclear fusion. The nuclear force between neutron-neutron, proton-neutron and proton-proton is approximately the same. The nuclear force does not depend on the electric charge. Nuclear force : At a nucleon separation of a femtometre - within a nucleus - there is a strong attractive force (104 N${10}^{4}\phantom{\rule{0.22em}{0ex}}N$). At equilibrium position : there is zero force acting for nucleons in a nucleus is at a separation of about 0.8 fm. The potential energy is minumim at r0=0.8 fm ${r}_{0}=0.8\phantom{\rule{0.22em}{0ex}}fm\phantom{\rule{0.22em}{0ex}}$ So a) the force is attractive for distances larger than 0.8 fm and b) the force is repulsive if they are separated by distances less than 0.8 fm. Fusion Reaction steps in Suni) 11H+11H→ 21H+ e++𝜈+0.42 MeV ${1}_{1}^{}H+{1}_{1}^{}H\to \phantom{\rule{0.44em}{0ex}}{1}_{2}^{}H+\phantom{\rule{0.22em}{0ex}}{e}^{+}+𝜈+0.42\phantom{\rule{0.22em}{0ex}}MeV\phantom{\rule{0.22em}{0ex}}$ii) e++e-→ 𝛾+ 𝛾+1.02 MeV ${e}^{+}+{e}^{-}\to \phantom{\rule{0.44em}{0ex}}𝛾+\phantom{\rule{0.22em}{0ex}}𝛾+1.02\phantom{\rule{0.22em}{0ex}}MeV\phantom{\rule{0.22em}{0ex}}$iii) 21H+11H→ 32He+ 𝛾+5.49 MeV ${1}_{2}^{}H+{1}_{1}^{}H\to \phantom{\rule{0.44em}{0ex}}{2}_{3}^{}He+\phantom{\rule{0.22em}{0ex}}𝛾+5.49\phantom{\rule{0.22em}{0ex}}MeV\phantom{\rule{0.22em}{0ex}}$iv) 32He+32He→ 42He+ 11H+11H+12.86 MeV ${2}_{3}^{}He+{2}_{3}^{}He\to \phantom{\rule{0.44em}{0ex}}{2}_{4}^{}He+\phantom{\rule{0.22em}{0ex}}{1}_{1}^{}H+{1}_{1}^{}H+12.86\phantom{\rule{0.22em}{0ex}}MeV\phantom{\rule{0.22em}{0ex}}$ So4 11H+2e-→42He+2e-+2𝜈+6𝛾+26.7 MeV $4\phantom{\rule{0.22em}{0ex}}{1}_{1}^{}H+2{e}^{-}\to {2}_{4}^{}He+2{e}^{-}+2𝜈+6𝛾+26.7\phantom{\rule{0.22em}{0ex}}MeV\phantom{\rule{0.22em}{0ex}}$ So, four hydrogen atoms combine to form an 42He${2}_{4}^{}He$ atom with a release of 26.7 MeV of energy. Mosley studied the characteristic X-ray spectrum of a number of a heavy elements and concluded that the spectra of different elements are very similar and with increasing atomic number, the spectral lines merely shift towards higher frequencies.𝜈=a(Z-b) $\sqrt{𝜈}=a\left(Z-b\right)\phantom{\rule{0.22em}{0ex}}$ where,Z = atomic numberb = screening constant or screening effect(Z - b) = effective nuclear charge Screening constant for K-series = 1Screening constant for L-series = 6.4 a and b are positive constant for one type of X-rays and for all elements. It is independent of Z
 Common Questions Q.1Show that 23892U${92}_{238}^{}U$ can not spontaneously emit a proton. Given: 23892U${92}_{238}^{}U$= 238.05079u, 23791Pa${91}_{237}^{}Pa$= 237.05121u 11He${1}_{1}^{}He$ = 1.00783u A.1If 23892U${92}_{238}^{}U$ spontaneously emit a proton, then the process will be 23892U→23791Pa +11He ${92}_{238}^{}U\to {91}_{237}^{}Pa\phantom{\rule{0.22em}{0ex}}+{1}_{1}^{}He\phantom{\rule{0.66em}{0ex}}$ Q value or disintegration energy=Q=(mU-mPa-mHe)c2$=Q=\left({m}_{U}-{m}_{Pa}-{m}_{He}\right){c}^{2}$⟹Q=(238.05079 u-237.05121 u-1.00783 u )c2$⟹Q=\left(238.05079\phantom{\rule{0.22em}{0ex}}u-237.05121\phantom{\rule{0.22em}{0ex}}u-1.00783\phantom{\rule{0.22em}{0ex}}u\phantom{\rule{0.22em}{0ex}}\right){c}^{2}$⟹Q=(-0.00825 u )c2$⟹Q=\left(-0.00825\phantom{\rule{0.22em}{0ex}}u\phantom{\rule{0.22em}{0ex}}\right){c}^{2}$ So here energy released is negative.So we need to add energy to make this possible.Therefore 23892U${92}_{238}^{}U$ can not spontaneously emit a proton. Q.2.Suggest an idea to convert a full wave bridge rectifier to a half wave rectifier by changing the connecting wire/s. Draw the diagram and explain your answer. A.2. Q-3Characteristic X-rays of frequency 4.2×1018 Hz $4.2×{10}^{18}\phantom{\rule{0.22em}{0ex}}Hz\phantom{\rule{0.22em}{0ex}}$ are produced when transitions from L - shell to K - shell take place in a certain target material. Use Moseley's law to determine the atomic number of the target material. (Rydberg's constant = 1.1×107 m-1$1.1×{10}^{7}\phantom{\rule{0.22em}{0ex}}{m}^{-1}$) A-3 We know: 𝜈=a(Z-b) $\sqrt{𝜈}=a\left(Z-b\right)\phantom{\rule{0.22em}{0ex}}$ 𝜈=CR(1n21-1n22) .(Z-b) $\sqrt{𝜈}=\sqrt{CR\left(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}}\right)}\phantom{\rule{0.44em}{0ex}}.\left(Z-b\right)\phantom{\rule{0.22em}{0ex}}$ where,Z = atomic numberb = screening constant or screening effect(Z - b) = effective nuclear charge Screening constant for K-series = 1Screening constant for L-series = 6.4 4.2×1018=3×108×1.1×107(112-122) .(Z-1) $\sqrt{4.2×{10}^{18}}=\sqrt{3×{10}^{8}×1.1×{10}^{7}\left(\frac{1}{{1}^{2}}-\frac{1}{{2}_{}^{2}}\right)}\phantom{\rule{0.44em}{0ex}}.\left(Z-1\right)\phantom{\rule{0.22em}{0ex}}$ ⟹ (Z-1)=41 $⟹\phantom{\rule{0.22em}{0ex}}\left(Z-1\right)=41\phantom{\rule{0.22em}{0ex}}$⟹ Z=42$⟹\phantom{\rule{0.22em}{0ex}}Z=42$ Q.If 178.5 pm is the wavelength of X-ray line of copper (atomic number 29) and 71 pm is the wavelength of the X-ray line of molybdenum (atomic number 42) then find the value of a and b in Moseley's equation A.5×107 Hz1/2 $5×{10}^{7}\phantom{\rule{0.22em}{0ex}}H{z}^{1/2}\phantom{\rule{0.22em}{0ex}}$ , b=1.4 Q.The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation A. the intensity increases.B. the minimum wavelength increases.C. the intensity remains unchanged.D. the minimum wavelength decreases. Answer: B Q.Assume that the nuclear binding energy per nucleon (B/A) versus mass number (A) is as shown in the figure. Use this plot to choose the correct choice(s) given below (A) Fusion of two nuclei with mass numbers lying in the range of 1 < A < 50$1\phantom{\rule{0.22em}{0ex}}<\phantom{\rule{0.22em}{0ex}}A\phantom{\rule{0.22em}{0ex}}<\phantom{\rule{0.22em}{0ex}}50$ will release energy (B) Fusion of two nuclei with mass numbers lying in the range of 51 < A < 100$51\phantom{\rule{0.22em}{0ex}}<\phantom{\rule{0.22em}{0ex}}A\phantom{\rule{0.22em}{0ex}}<\phantom{\rule{0.22em}{0ex}}100$ will release energy (C) Fission of a nucleus lying in the mass range of 100 < A < 200$100\phantom{\rule{0.22em}{0ex}}<\phantom{\rule{0.22em}{0ex}}A\phantom{\rule{0.22em}{0ex}}<\phantom{\rule{0.22em}{0ex}}200$ will release energy when broken into two equal fragments (D) Fission of a nucleus lying in the mass range of 200 < A < 260$200\phantom{\rule{0.22em}{0ex}}<\phantom{\rule{0.22em}{0ex}}A\phantom{\rule{0.22em}{0ex}}<\phantom{\rule{0.22em}{0ex}}260$ will release energy when broken into two equal fragments Answer: B, D
 NCERT Questions Q.13.3 Obtain the binding energy (in MeV) of a nitrogen nucleus (147N${7}_{14}^{}N$) given , m of (147N${7}_{14}^{}N$) =14.00307 uA.13.3Number of protons in Nitrogen=7 Number of neutrons in Nitrogen=7 Total mass of protons=7×1.007825$=7×1.007825$ uMass of neutrons=7×1.00866$=7×1.00866$5 u Total expected mass of nitrogen= 14.11543 u𝛥m=14.11543-14.00307=0.11236 u $𝛥m=14.11543-14.00307=0.11236\phantom{\rule{0.22em}{0ex}}u\phantom{\rule{0.22em}{0ex}}$1 u = 931.5 MeV/c2$1\phantom{\rule{0.22em}{0ex}}u\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}931.5\phantom{\rule{0.22em}{0ex}}MeV/{c}^{2}$Binding energy=0.11236×931.5=104.66334 MeV$Binding\phantom{\rule{0.22em}{0ex}}energy=0.11236×931.5=104.66334\phantom{\rule{0.22em}{0ex}}MeV$ Q.13.9Obtain the amount of 6027Co ${27}_{60}^{}Co\phantom{\rule{0.22em}{0ex}}$ necessary to provide a radioactive source of 8.0 mCi strength. The half-life of 6027Co${27}_{60}^{}Co$ is 5.3 years.A.13.9T1/2=0.693𝜆 ${T}_{1/2}=\frac{0.693}{𝜆}\phantom{\rule{0.22em}{0ex}}$ T1/2=${T}_{1/2}=$5.3 years=5.3 x 365 x 24 x 3600 = 167140800 secondsDisintegration constant=𝜆=0.693T1/2 =0.693167140800=4.1462×10-9 $Disintegration\phantom{\rule{0.22em}{0ex}}constant=𝜆=\frac{0.693}{{T}_{1/2}}\phantom{\rule{0.22em}{0ex}}=\frac{0.693}{167140800}=4.1462×{10}^{-9}\phantom{\rule{0.44em}{0ex}}$ 1 curie = 1 Ci = 3.7×1010 Bq (decays per second)$1\phantom{\rule{0.22em}{0ex}}curie\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}1\phantom{\rule{0.22em}{0ex}}Ci\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}3.7×{10}^{10}\phantom{\rule{0.22em}{0ex}}Bq\phantom{\rule{0.22em}{0ex}}\left(decays\phantom{\rule{0.22em}{0ex}}per\phantom{\rule{0.22em}{0ex}}second\right)$R=8.0 mCi = 2.96×108 Bq (decays per second)$R=8.0\phantom{\rule{0.22em}{0ex}}mCi\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}2.96×{10}^{8}\phantom{\rule{0.22em}{0ex}}Bq\phantom{\rule{0.22em}{0ex}}\left(decays\phantom{\rule{0.22em}{0ex}}per\phantom{\rule{0.22em}{0ex}}second\right)$ R=𝜆N $R=𝜆N\phantom{\rule{0.22em}{0ex}}$⟹2.96×108=4.1462×10-9×N $⟹2.96×{10}^{8}=4.1462×{10}^{-9}×N\phantom{\rule{0.44em}{0ex}}$⟹N=7.139×1016 atoms $⟹N=7.139×{10}^{16}\phantom{\rule{0.22em}{0ex}}atoms\phantom{\rule{0.22em}{0ex}}$ So, mass of 6027Co=7.139×10166.023×1023×60=7.11×10-6 gm $So,\phantom{\rule{0.22em}{0ex}}mass\phantom{\rule{0.22em}{0ex}}of\phantom{\rule{0.22em}{0ex}}{27}_{60}^{}Co=\frac{7.139×{10}^{16}}{6.023×{10}^{23}}×60=7.11×{10}^{-6}\phantom{\rule{0.22em}{0ex}}gm\phantom{\rule{0.22em}{0ex}}$ Q.13.10 The half-life of 9038Sr $\phantom{\rule{0.44em}{0ex}}{38}_{90}^{}Sr\phantom{\rule{0.22em}{0ex}}$ is 28 years. What is the disintegration rate of15 mg of this isotope? A.13.10 T1/2=${T}_{1/2}=$28 years=883008000 secondsT1/2=8.83008×108 =0.693𝜆 ${T}_{1/2}=8.83008×{10}^{8}\phantom{\rule{0.22em}{0ex}}=\frac{0.693}{𝜆}\phantom{\rule{0.22em}{0ex}}$𝜆=7.8482×10-10 $𝜆=7.8482×{10}^{-10}\phantom{\rule{0.22em}{0ex}}$15mg=0.015 g$15mg=0.015\phantom{\rule{0.22em}{0ex}}g$number of nuclides of Sr=6.023×1023×0.01590=1.004×1020 $=\frac{6.023×{10}^{23}×0.015}{90}=1.004×{10}^{20}\phantom{\rule{0.22em}{0ex}}$ R=𝜆N=7.8482×10-10 ×1.004×1020 =7.879×1010Bq $R=𝜆N=7.8482×{10}^{-10}\phantom{\rule{0.22em}{0ex}}×1.004×{10}^{20}\phantom{\rule{0.22em}{0ex}}=7.879×{10}^{10}Bq\phantom{\rule{0.44em}{0ex}}$1 curie = 1 Ci = 3.7×1010 Bq (decays per second)$1\phantom{\rule{0.22em}{0ex}}curie\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}1\phantom{\rule{0.22em}{0ex}}Ci\phantom{\rule{0.22em}{0ex}}=\phantom{\rule{0.22em}{0ex}}3.7×{10}^{10}\phantom{\rule{0.22em}{0ex}}Bq\phantom{\rule{0.22em}{0ex}}\left(decays\phantom{\rule{0.22em}{0ex}}per\phantom{\rule{0.22em}{0ex}}second\right)$ R=2.13 Ci $R=2.13\phantom{\rule{0.22em}{0ex}}Ci\phantom{\rule{0.44em}{0ex}}$ Q.13.11 Obtain approximately the ratio of the nuclear radii of the gold isotope 19779Au$\phantom{\rule{0.22em}{0ex}}{79}_{197}^{}Au$ and the silver isotope 10747Ag$\phantom{\rule{0.22em}{0ex}}{47}_{107}^{}Ag$ A.13.11 Radius of Nucleus of mass number AR=R0A13$R={R}_{0}{A}^{\frac{1}{3}}$R0=1.2×10-15 m ${R}_{0}=1.2×{10}^{-15}\phantom{\rule{0.22em}{0ex}}m\phantom{\rule{0.22em}{0ex}}$ RAuRA6=(AAuAA6)13=(197107)13 $\frac{{R}_{Au}}{{R}_{A6}}=\left(\frac{{A}_{Au}}{{A}_{A6}}{\right)}^{\frac{1}{3}}=\left(\frac{197}{107}{\right)}^{\frac{1}{3}}\phantom{\rule{0.22em}{0ex}}$RAuRA6=1.2256 $\frac{{R}_{Au}}{{R}_{A6}}=1.2256\phantom{\rule{0.22em}{0ex}}$ Q.13.19 How long can an electric lamp of 100W be kept glowing by fusion of2.0 kg of deuterium? Take the fusion reaction as21H+21H→42He+n+3.27 MeV ${1}_{2}^{}H+{1}_{2}^{}H\to {2}_{4}^{}He+n+3.27\phantom{\rule{0.22em}{0ex}}MeV\phantom{\rule{0.22em}{0ex}}$ A.13.19 3.27 MeV=3.27×106×1.6×10-19 Joules$3.27\phantom{\rule{0.22em}{0ex}}MeV=3.27×{10}^{6}×1.6×{10}^{-19}\phantom{\rule{0.22em}{0ex}}Joules$2kg deuterium⟹6.023×1023×1000 nuclides=6.023×1026 nuclides $2kg\phantom{\rule{0.22em}{0ex}}deuterium⟹6.023×{10}^{23}×1000\phantom{\rule{0.22em}{0ex}}nuclides=6.023×{10}^{26}\phantom{\rule{0.22em}{0ex}}nuclides\phantom{\rule{0.22em}{0ex}}$2kg deuterium⟹6.023×1026 nuclides $2kg\phantom{\rule{0.22em}{0ex}}deuterium⟹6.023×{10}^{26}\phantom{\rule{0.22em}{0ex}}nuclides\phantom{\rule{0.22em}{0ex}}$ Total energy released=3.27×106×1.6×10-19×6.023×10262 $=\frac{3.27×{10}^{6}×1.6×{10}^{-19}×6.023×{10}^{26}}{2}\phantom{\rule{0.22em}{0ex}}$Total energy released=15.756168×1013 Joules$=15.756168×{10}^{13}\phantom{\rule{0.22em}{0ex}}Joules$Total time lamp will glow=15.756168×1013100=15.756168×1011 seconds$=\frac{15.756168×{10}^{13}}{100}=15.756168×{10}^{11}\phantom{\rule{0.22em}{0ex}}seconds$Total time lamp will glow=15.756168×10113600×24×365=4.99×104 years$=\frac{15.756168×{10}^{11}}{3600×24×365}=4.99×{10}^{4}\phantom{\rule{0.22em}{0ex}}years$ Q.13.20 Calculate and compare the energy released by a) fusion of 1.0 kg of hydrogen deep within Sun and b) the fission of 1.0 kg of 235U in a fission reactor.A.13.20 (a)Number of atoms of hydrogen in 1 kg =1000×6.023×1023$=1000×6.023×{10}^{23}$We know that four hydrogen atoms combine to form an 42He${2}_{4}^{}He$ atom with a release of 26.7 MeV of energy.So total energy released E=14×1000×6.023×1023×26.7 MeV$\phantom{\rule{0.22em}{0ex}}E=\frac{1}{4}×1000×6.023×{10}^{23}×26.7\phantom{\rule{0.22em}{0ex}}MeV$ E=40.2×1026 MeV$\phantom{\rule{0.22em}{0ex}}E=40.2×{10}^{26}\phantom{\rule{0.22em}{0ex}}MeV$ (b)In a single uranium (235U) fission 200 MeV is released.Number of atoms of uranium (235U) in 1 kg =1000235×6.023×1023$=\frac{1000}{235}×6.023×{10}^{23}$ So total energy released E=1000235×6.023×1023×200 MeV$\phantom{\rule{0.22em}{0ex}}E=\frac{1000}{235}×6.023×{10}^{23}×200\phantom{\rule{0.22em}{0ex}}MeV$ E=5.12×1026 MeV$\phantom{\rule{0.22em}{0ex}}E=5.12×{10}^{26}\phantom{\rule{0.22em}{0ex}}MeV$