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Q.Write the nuclear reaction of 92235U^{235}_{92}U bombarded with a slow neutron. Calculate the energy released in KWh in the fission of 50 kg of 92235U^{235}_{92}U. OR Using Bohr's quantization condition for angular momentum of an electron revolving around the hydrogen nucleus, establish the expression of the radius of the stationary orbits of hydrogen atom.

Manipur CohsemCOHSEM Manipur Higher Secondary Board 2019Subjective· 5mImportance★★★★★
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Each U-235 fission releases ~200 MeV; multiplying by the number of atoms in 50 kg and converting to kWh gives about 1.14×1091.14\times10^9 kWh. (OR alternative: Bohr's quantization gives rn=n2h2ε0πme2r_n=\dfrac{n^2h^2\varepsilon_0}{\pi m e^2}.)

Part 1 (primary): Nuclear fission of U-235 and energy released

Nuclear reaction. When 92235U^{235}_{92}U is bombarded with a slow ("thermal") neutron, it forms an unstable compound nucleus 92236U∗^{236}_{92}U^*, which splits (fissions) into two lighter nuclei, releasing 2–3 fast neutrons and a large amount of energy. A representative fission reaction is:

92235U+01n  ⟶  56141Ba+3692Kr+3 01n+Q^{235}_{92}U + {}^1_0n \;\longrightarrow\; {}^{141}_{56}Ba + {}^{92}_{36}Kr + 3\,{}^1_0n + Q

(Many different fission fragment pairs are possible; this is one typical example. Charge and mass number both balance: 92+0=56+36+092+0=56+36+0; 235+1=141+92+3235+1=141+92+3.)

Energy per fission. Each fission event of 92235U^{235}_{92}U releases, on average, about

Q≈200 MeV=200×1.6×10−13 J=3.2×10−11 JQ \approx 200\,MeV = 200\times1.6\times10^{-13}\,J = 3.2\times10^{-11}\,J

Number of atoms in 50 kg of 92235U^{235}_{92}U. Using Avogadro's number NA=6.023×1023N_A=6.023\times10^{23} per mole and molar mass 235 g/mol235\,g/mol:

N=mM×NA=50×103 g235 g/mol×6.023×1023≈212.77×6.023×1023≈1.28×1026 atomsN = \dfrac{m}{M}\times N_A = \dfrac{50\times10^3\,g}{235\,g/mol}\times6.023\times10^{23} \approx 212.77 \times 6.023\times10^{23} \approx 1.28\times10^{26}\ \text{atoms}

Total energy released:

E=N×Q=1.28×1026×3.2×10−11≈4.10×1015 JE = N\times Q = 1.28\times10^{26}\times3.2\times10^{-11} \approx 4.10\times10^{15}\,J

Converting to kWh (using 1 kWh=3.6×106 J1\,kWh = 3.6\times10^6\,J):

E=4.10×10153.6×106≈1.14×109 kWhE = \dfrac{4.10\times10^{15}}{3.6\times10^6} \approx 1.14\times10^{9}\,kWh

Part 2 (OR): Bohr's radius of the hydrogen atom's stationary orbits

Bohr postulated that an electron can revolve only in those orbits where its angular momentum is quantised in integral multiples of h/2πh/2\pi:

mvr=nh2π,n=1,2,3,…...(1)mvr = \dfrac{nh}{2\pi}, \quad n=1,2,3,\ldots \quad\text{...(1)}

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