Questions: Nuclei

Q1. Find the binding energy of a nitrogen-14 nucleus, given m(14N) = 14.00307 u.

Answer (English): Nitrogen-14 has Z = 7 and A = 14, so it has 7 protons and 7 neutrons. The mass of the parts is 7 × 1.007825 + 7 × 1.008665 = 14.11543 u, so the mass defect is 14.11543 - 14.00307 = 0.11236 u. Multiplying by 931.5 MeV per u gives a binding energy of about 104.7 MeV. Dividing by 14 gives about 7.48 MeV per nucleon.

Hinglish: Nitrogen-14 mein 7 protons aur 7 neutrons hain. Parts ka mass 7 × 1.007825 + 7 × 1.008665 = 14.11543 u, isliye mass defect 0.11236 u. 931.5 MeV per u se binding energy ≈ 104.7 MeV, aur per nucleon ≈ 7.48 MeV.

Q2. Obtain the binding energy of iron-56 and bismuth-209, given m(56Fe) = 55.934939 u and m(209Bi) = 208.980388 u.

Answer (English): Iron-56 has Z = 26 and 30 neutrons, so the parts mass 26 × 1.007825 + 30 × 1.008665 = 56.4634 u. The defect is 0.528461 u, giving about 492.3 MeV, or about 8.79 MeV per nucleon. Bismuth-209 has Z = 83 and 126 neutrons, parts mass 210.7413 u, defect 1.760877 u, binding energy about 1640 MeV, or about 7.85 MeV per nucleon. Iron is more tightly bound per nucleon than bismuth, which is why heavy nuclei can release energy by fission.

Hinglish: Fe-56 mein 26 protons aur 30 neutrons hain, parts ka mass 56.4634 u, defect 0.528461 u, BE ≈ 492.3 MeV aur per nucleon ≈ 8.79 MeV. Bi-209 mein 83 protons aur 126 neutrons hain, defect 1.760877 u, BE ≈ 1640 MeV aur per nucleon ≈ 7.85 MeV. Iron bismuth se zyada tightly bound hai, isliye heavy nuclei fission se energy de sakte hain.

Q3. A coin of mass 3.0 g is assumed to be made entirely of copper-63 of mass 62.92960 u. Find the energy needed to separate all its nucleons.

Answer (English): Copper-63 has Z = 29 and 34 neutrons. The parts mass 29 × 1.007825 + 34 × 1.008665 = 63.52154 u, so the defect per atom is 0.591935 u and the binding energy per atom is about 551.4 MeV. The number of atoms in 3.0 g is (3.0/63) × 6.023 × 10 to the 23, about 2.87 × 10 to the 22. The total energy is therefore about 1.58 × 10 to the 25 MeV, which is about 2.5 × 10 to the 12 J.

Hinglish: Cu-63 mein 29 protons aur 34 neutrons hain; per atom defect 0.591935 u aur BE ≈ 551.4 MeV. 3.0 g mein atoms = (3/63) × 6.023×10²³ ≈ 2.87×10²². Total energy ≈ 1.58×10²⁵ MeV ≈ 2.5×10¹² J.

Q4. Find the ratio of the nuclear radii of gold-197 and silver-107.

Answer (English): The radius goes as A to the power one-third, so the ratio is (197/107) to the power one-third, which is about 1.23. Gold is the larger nucleus, but only by about a quarter, because the cube-root dependence compresses a large change in mass number into a small change in radius.

Hinglish: Radius A^(1/3) ke saath badhta hai, isliye ratio = (197/107)^(1/3) ≈ 1.23. Gold bada hai lekin sirf lagbhag ek chauthai, kyunki cube root ki wajah se mass number ka bada farq radius mein chhota dikhta hai.

Q5. Find the Q value of (i) 2H + 2H going to 3H + 1H and (ii) 12C + 12C going to 20Ne + 4He, using the masses given in the exercise.

Answer (English): (i) Initial mass is 2 × 2.014102 = 4.028204 u and final mass is 3.016049 + 1.007825 = 4.023874 u, so the defect is 0.00433 u and Q is about 4.03 MeV, positive, so the reaction is exothermic. (ii) Initial mass is 24.000000 u and final mass is 19.992439 + 4.002603 = 23.995042 u, so Q is about 4.62 MeV, again positive. Both reactions release energy.

Hinglish: (i) Initial 4.028204 u, final 4.023874 u, defect 0.00433 u, Q ≈ 4.03 MeV (exothermic). (ii) Initial 24.000000 u, final 23.995042 u, Q ≈ 4.62 MeV, yeh bhi exothermic hai.

Q6. Is the fission of iron-56 into two aluminium-28 nuclei energetically possible? Take m(56Fe) = 55.93494 u and m(28Al) = 27.98191 u.

Answer (English): The initial mass is 55.93494 u and the final mass is 2 × 27.98191 = 55.96382 u. The mass increases by 0.02888 u, so Q is negative, about -26.9 MeV. The reaction cannot happen on its own: iron-56 sits near the maximum of the binding energy curve, so splitting it into two lighter nuclei makes the nucleons less tightly bound, which needs energy instead of releasing it.

Hinglish: Initial 55.93494 u, final 55.96382 u, mass 0.02888 u badh jaata hai, isliye Q ≈ −26.9 MeV. Reaction apne aap nahi ho sakta: Fe-56 binding energy curve ke maximum ke paas hai, isliye todne par nucleons kam tightly bound hote hain aur energy deni padti hai.

Q7. The average energy released per fission of plutonium-239 is 180 MeV. Find the energy released when all the atoms in 1 kg of pure plutonium-239 fission.

Answer (English): One kilogram contains 1000/239 moles of nuclei, which is 4.184 moles, and multiplying by Avogadro's number gives about 2.52 × 10 to the 24 nuclei. Multiplying by 180 MeV per fission gives about 4.54 × 10 to the 26 MeV, which is about 7.3 × 10 to the 13 J. That is the order of energy a large power station produces in a day.

Hinglish: 1 kg mein 1000/239 = 4.184 mole nuclei, yaani lagbhag 2.52×10²⁴ nuclei. Har fission 180 MeV deta hai, isliye total ≈ 4.54×10²⁶ MeV ≈ 7.3×10¹³ J, jo ek bade power station ke ek din ke energy ke order ka hai.

Q8. A 100 W lamp is run on the fusion of 2.0 kg of deuterium through the reaction shown. For how long can it glow?

Answer (English): 2.0 kg of deuterium contains 1000 moles of nuclei, or 6.023 × 10 to the 26 nuclei. Each reaction uses two deuterium nuclei, so the number of reactions is about 3.01 × 10 to the 26 and the energy released is 3.27 MeV times that, about 9.85 × 10 to the 26 MeV, or 1.58 × 10 to the 14 J. A 100 W lamp uses 100 J per second, so it could glow for about 1.58 × 10 to the 12 s, which is about 5 × 10 to the 4 years.

Hinglish: 2.0 kg deuterium mein 1000 mole nuclei, yaani 6.023×10²⁶ nuclei. Har reaction do nuclei leta hai, isliye reactions ≈ 3.01×10²⁶ aur energy ≈ 9.85×10²⁶ MeV ≈ 1.58×10¹⁴ J. 100 W lamp 100 J per second leta hai, isliye time ≈ 1.58×10¹² s ≈ 5×10⁴ years.

Q9. Two deuterons of radius 2.0 fm meet head on. Work out the height of the Coulomb barrier between them.

Answer (English): At the moment of touching, the centres are 4.0 fm apart, so the electrical potential energy is (1/4πε0) e squared/(4.0 fm). Substituting the constants gives about 5.8 × 10 to the -14 J, which is about 0.36 MeV. This is the energy each deuteron pair must have to overcome the barrier, which is why fusion needs very high temperature.

Hinglish: Touch karte waqt centres 4.0 fm door hote hain, isliye potential energy (1/4πε0)e²/(4.0 fm) ≈ 5.8×10⁻¹⁴ J ≈ 0.36 MeV. Yahi barrier ko paar karne ki energy hai, isliye fusion ke liye bahut high temperature chahiye.

Q10. Starting from R = R0A^(1/3), show that nuclear matter density is nearly constant.

Answer (English): The volume of a nucleus is (4/3)πR cubed = (4/3)πR0 cubed A. If the mass of each nucleon is m, the mass of the nucleus is approximately mA. Density is mass divided by volume, so it equals 3mA/(4πR0 cubed A) = 3m/(4πR0 cubed). The mass number A cancels, leaving a constant that depends only on the nucleon mass and R0. Numerically it is of the order of 10 to the 17 kg per cubic metre for every nucleus.

Hinglish: Volume = (4/3)πR³ = (4/3)πR0³A. Nucleus ka mass ≈ mA, isliye density = mA/volume = 3m/(4πR0³). A cancel ho jaata hai, aur density sirf nucleon mass aur R0 par depend karti hai, yaani har nucleus ke liye lagbhag 10¹⁷ kg/m³.

To check how much of this stuck, take the five-question quiz below. Each answer explains the reasoning, so read the explanation even when you get it right.

Quiz: the five question check for this chapter is set from the same topics. Attempt it after the questions above, then revisit any topic page where you lost a mark.