Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Atoms and Nuclei question

2020 · 2 Sep · Shift 1 · Q40
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Atoms and Nuclei
  5. /2020 · 2 Sep · Shift 1 · Q40

Atoms and Nuclei question

2020 · 2 Sep · Shift 1 · Q40

JEE MainPhysicsAtoms and NucleiMCQ+4 / −1
In a reactor, 2 kg of 92U235 fuel is fully used up in 30 days. The energy released per fission is 200 MeV. Given that the Avogadro number, N = 6.023 ×\times× 1026 per kilo mole and 1 eV = 1.6 × 10–19 J. The power output of the reactor is close to
  1. A
    125 MW
  2. B
    60 MW
  3. C
    54 MW
  4. D
    35 MW
View written solutionFree

Correct answer: B

  1. Find number of 235U^{235}\text{U}235U nuclei in 2 kg2\,\text{kg}2kg

The molar mass of 235U^{235}\text{U}235U is 235 kg per kilo-mole235\,\text{kg per kilo-mole}235kg per kilo-mole.

So, number of kilo-moles in 2 kg2\,\text{kg}2kg is

2235 kmol\frac{2}{235}\,\text{kmol}2352​kmol

Hence total number of nuclei is

N0=2235×6.023×1026N_0=\frac{2}{235}\times 6.023\times 10^{26}N0​=2352​×6.023×1026 N0≈5.126×1024N_0\approx 5.126\times 10^{24}N0​≈5.126×1024
  1. Energy released per fission in joules

Given:

200 MeV=200×106 eV200\,\text{MeV} = 200\times 10^6\,\text{eV}200MeV=200×106eV

Using

1 eV=1.6×10−19 J1\,\text{eV}=1.6\times 10^{-19}\,\text{J}1eV=1.6×10−19J

Therefore,

Ef=200×106×1.6×10−19E_f = 200\times 10^6\times 1.6\times 10^{-19}Ef​=200×106×1.6×10−19 Ef=3.2×10−11 JE_f = 3.2\times 10^{-11}\,\text{J}Ef​=3.2×10−11J
  1. Total energy released
E=N0EfE=N_0E_fE=N0​Ef​ E=(5.126×1024)(3.2×10−11)E=(5.126\times 10^{24})(3.2\times 10^{-11})E=(5.126×1024)(3.2×10−11) E≈1.64×1014 JE\approx 1.64\times 10^{14}\,\text{J}E≈1.64×1014J
  1. Time for consumption of fuel

Given 303030 days:

t=30×24×3600t=30\times 24\times 3600t=30×24×3600 t=2.592×106 st=2.592\times 10^6\,\text{s}t=2.592×106s
  1. Power output
P=EtP=\frac{E}{t}P=tE​ P=1.64×10142.592×106P=\frac{1.64\times 10^{14}}{2.592\times 10^6}P=2.592×1061.64×1014​ P≈6.33×107 WP\approx 6.33\times 10^7\,\text{W}P≈6.33×107W P≈63 MWP\approx 63\,\text{MW}P≈63MW

This is closest to 60 MW60\,\text{MW}60MW.

  1. Option check
  • A: 125 MW125\,\text{MW}125MW — too large
  • B: 60 MW60\,\text{MW}60MW — closest
  • C: 54 MW54\,\text{MW}54MW — somewhat close, but not best
  • D: 35 MW35\,\text{MW}35MW — too small

Therefore, the correct option is B.

PreviousNext

More from Atoms and Nuclei

  • In a hydrogen atom the electron makes a transition from (n + 1)th level to the nth level. If n >> 1, the frequency of radiation emitted is proportional to :2020 · MCQ
  • Hydrogen ion and singly ionized helium atom are accelerated, from rest, through the same potential difference. The ratio of final speeds of hydrogen and helium ions is close to :2020 · MCQ
  • The radius R of a nucleus of mass number A can be estimated by the formula R = (1.3 × 10–15)A1/3 m. It follows that the mass density of a nucleus is of the order of : (Mprot. ≅ Mneut ≃ 1.67 × 10–27 kg)2020 · MCQ
  • In the line spectra of hydrogen atoms, difference between the largest and the shortest wavelengths of the Lyman series is 304 A0​. The corresponding difference for the Paschan series in A0​ is : ​…2020 · Numerical
  • Find the Binding energy per neucleon for 50120​Sn. Mass of proton mp = 1.00783 U, mass of neutron mn = 1.00867 U and mass of tin nucleus mSn = 119.902199 U. (take 1U = 931 MeV)2020 · MCQ
  • A particle of mass 200 MeV/c2 collides with a hydrogen atom at rest. Soon after the collision the particle comes to rest, and the atom recoils and goes to its first excited state. The initial kinetic energy of the particle (in eV) is 4N​…2020 · Numerical
  • You are given that Mass of 37​Li= 7.0160u, Mass of 24​He= 4.0026u and Mass of 11​H= 1.0079u. When 20 g of 37​Li is converted into 24​He by proton capture, the energy liberated, (in kWh), is : [Mass of nucleon = 1…2020 · MCQ
  • Given the masses of various atomic particles mp = 1.0072 u, mn = 1.0087 u, me = 0.000548 u, mv​= 0, md = 2.0141 u, where p ≡ proton, n ≡ neutron, e ≡ electron, v≡ antineutrino and d ≡…2020 · MCQ