Q.48. The energy released when kg of is converted into by proton bombardment is eV. The value of is ________ (Nearest integer)
(Mass of u, mass of u, mass of proton u and 1 u MeV/c and Avogad

Question

Q.48. The energy released when  kg of  is converted into  by proton bombardment is  eV. The value of  is ________ (Nearest integer)
(Mass of  u, mass of  u, mass of proton  u and 1 u  MeV/c and Avogadro number )

FirstInTest · Accepted Answer

Correct answer: 6. The nuclear reaction is $^7Li + ^1H ightarrow 2 ^4He$. First, calculate the mass defect ($\Delta m$). Mass of reactants = mass($^7Li$) + mass($^1H$) = $7.0183 	ext{ u} + 1.008 	ext{ u} = 8.0263 	ext{ u}$. Mass of products = $2 	imes$ mass($^4He$) = $2 	imes 4.004 	ext{ u} = 8.008 	ext{ u}$. Mass defect $\Delta m = (8.0263 - 8.008) 	ext{ u} = 0.0183 	ext{ u}$. Energy released per reaction $E_{reaction} = \Delta m 	imes 931 	ext{ MeV/c}^2 = 0.0183 	imes 931 	ext{ MeV} = 17.0373 	ext{ MeV}$. Convert to eV: $17.0373 	imes 10^6 	ext{ eV}$. Now, calculate the number of moles of $^7Li$ in $\frac{7}{17.13}$ kg. Molar mass of $^7Li$ is approximately 7 g/mol. Number of moles of $^7Li = \frac{	ext{mass}}{	ext{molar mass}} = \frac{7/17.13 	ext{ kg}}{7 	ext{ g/mol}} = \frac{7000/17.13 	ext{ g}}{7 	ext{ g/mol}} = \frac{1000}{17.13}$ mol. Number of $^7Li$ atoms = Number of moles $	imes N_A = \frac{1000}{17.13} 	imes 6.0 	imes 10^{23}$. Since each $^7Li$ atom undergoes one reaction, this is the total number of reactions. Total energy released $E_{total} = E_{reaction} 	imes 	ext{Number of atoms} = (17.0373 	imes 10^6 	ext{ eV}) 	imes (\frac{1000}{17.13} 	imes 6.0 	imes 10^{23})$. $E_{total} = \frac{17.0373 	imes 10^6 	imes 1000 	imes 6.0 	imes 10^{23}}{17.13} 	ext{ eV}$. $E_{total} = \frac{17.0373 	imes 6.0}{17.13} 	imes 10^{32} 	ext{ eV}$. $\frac{17.0373 	imes 6.0}{17.13} \approx \frac{102.2238}{17.13} \approx 5.967 \approx 6$. So, $\alpha = 6$. The solution uses $\Delta m = (7.018 + 1.008) - 2(4.004) = 0.0183$ u. This is correct. Energy released per reaction $E = \Delta m c^2 = 0.0183 	imes 931$ MeV. Number of moles of Li = $\frac{7}{17.13 	ext{ kg}} / (7 	ext{ g/mol}) = \frac{7 	imes 10^3 	ext{ g}}{17.13 	imes 7 	ext{ g/mol}} = \frac{10^3}{17.13}$ mol. Number of molecules = $\frac{10^3}{17.13} 	imes 6 	imes 10^{23}$. Total energy released $E_{Total} = 0.0183 	imes 931 	imes \frac{10^3}{17.13} 	imes 6 	imes 10^{23}$ MeV. To get it in eV, multiply by $10^6$. $E_{Total} = 0.0183 	imes 931 	imes \frac{10^3}{17.13} 	imes 6 	imes 10^{23} 	imes 10^6 	ext{ eV} = \frac{0.0183 	imes 931 	imes 6}{17.13} 	imes 10^{32} 	ext{ eV}$. $\frac{0.0183 	imes 931 	imes 6}{17.13} = \frac{17.0373 	imes 6}{17.13} = \frac{102.2238}{17.13} \approx 5.967 \approx 6$. So $\alpha = 6$.

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