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: $m(n)=1.0087 amu, M(^{235}U)=235.0439 amu, M(^{99}Zr)=98.916 amu M(^{134}Te)=133.9115 amu$

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Solution

Q value is equal to mass-energy equivalence
$Q = △ m c^{2} △ m = m_{(n)} + m_{(^{235} U)} - m_{(^{99} Z e)} - m_{(^{134} T e)} - 3 m_{(n)} = m_{(^{235} U)} - m_{(^{99} Z e)} - m_{(^{134} T e)} - 2 m_{(n)} = 235.0439 - 98.916 - 133.9115 - 2 \times 1.0087 = 0.199 a m u Q = 0.199 \times 931 M e V Q = 185.269 M e V$

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Similar questions

^{235} U

^{238} U

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

_{92}^{235} U + N \to_{38}^{94} Sr +_{54}^{140} Xe + 2 n .

β

_{94}^{38} Sr

_{94}^{40} Zr

_{140}^{54} Xe

_{140}^{58} Ce .

m (^{235} U) = 235.439 u, m (n) = 1.00866 u, m (^{94} Z r) = 93.9064 u, m (^{140} C e) = 139.9055 u, 1 u = 931 M e V]

1.40 \times 10^{- 10} m

(3 / 2) k T

300 K

m_{n} = 1.6 \times 10^{- 27} k g)

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