Calculate the excitation energy of the compound nuclei produced when $^{235} U$ and $^{238} U$ absorb thermal neutrons. Given:
$M (^{235} U) = 235.0439 a m u$ $M (n) = 1.0087 a m u$ $M (^{238} U) = 238.0508 a m u$ $M (^{236} U) = 236.0456 a m u$ $M (^{239} U) = 239.0543 a m u$

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Solution

$^{235} U + n \to^{236} U △ m = - 236.0456 + 1.0087 235.0439 = 0.007 E = △ m c^{2} = 0.007 \times 931 M e V = 6.517$

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Similar questions

Q. Find the amount of energy released in

kWh

when

1 g

of Uranium atoms

_{92} U^{235} (235.0439

amu)

undergoes fission by slow neutron

(1.0087 amu)

and is split into Krypton

_{36} K r^{92} (91.8973 amu)

and Barium

_{56} B r^{141} (140.9139 amu)

(assuming no energy is lost).

[1 amu c^{2} = 931.5 MeV]

Q. Calculate the ground state Q value of the induced fission reaction in the equation

n +_{92}^{235} U \to_{92}^{236} U^{*} \to_{40}^{99} Z e +_{52}^{134} T e + 3 n

if the neutron is thermal. A thermal neutron is in the thermal equilibrium with its environment; it has an average kinetic energy given by (3/2) kT. Given: $m(n)=1.0087 amu, M(^{235}U)=235.0439 amu, M(^{99}Zr)=98.916 amu M(^{134}Te)=133.9115 amu$

The masses of neutron and proton are 1.0087 amu and 1.0073 amu respectively. If the neutrons and protons combine to form a helium nucleus (alpha particles) of mass 4.0015 amu. The binding energy of the helium nucleus will be [1 amu= 931 MeV]

Q. (a) Calculate the energy released if ²³⁸U emits an α-particle. (b) Calculate the energy to be supplied to ²³⁸U it two protons and two neutrons are to be emitted one by one. The atomic masses of ²³⁸U, ²³⁴Th and ⁴He are 238.0508 u, 234.04363 u and 4.00260 u respectively.

Q. Calculate the binding energy per nucleon for

_{10}^{20} N e,_{26}^{56} F e

and

_{92}^{238} U

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_{10}^{20} N e

is 19.9924 amu, mass of

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is 55.93492 amu,

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is 238.050783 amu

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Calculate the excitation energy of the compound nuclei produced when 235U and 238U absorb thermal neutrons. Given:M(235U)=235.0439amuM(n)=1.0087amuM(238U)=238.0508amuM(236U)=236.0456amuM(239U)=239.0543amu

235U+n→236U△m=−236.0456+1.0087235.0439=0.007E=△mc2=0.007×931MeV=6.517

Calculate the excitation energy of the compound nuclei produced when $^{235} U$ and $^{238} U$ absorb thermal neutrons. Given:
$M (^{235} U) = 235.0439 a m u$ $M (n) = 1.0087 a m u$ $M (^{238} U) = 238.0508 a m u$ $M (^{236} U) = 236.0456 a m u$ $M (^{239} U) = 239.0543 a m u$

$^{235} U + n \to^{236} U △ m = - 236.0456 + 1.0087 235.0439 = 0.007 E = △ m c^{2} = 0.007 \times 931 M e V = 6.517$