The disintegration energy Q for the nuclear fission of U 235 ? C 140 e + Z 94 r + n is ___ MeV. (Given atomic masses of U 235 : 235 . 0439 u ; C 140 e : 139 . 9054 u ; Z 94 r : 93 . 9063 u ; n : 1 . 0086 u , and value of c 2 = 931 MeV / u .)

Question

The disintegration energy $Q$ for the nuclear fission of ${}^{235}U \to {}^{140}C e + {}^{94}Z r + n$ is ___ MeV.

(Given atomic masses of ${}^{235}U : 235.0439 u; {}^{140}C e : 139.9054 u; {}^{94}Z r : 93.9063 u; n : 1.0086 u$ , and value of $c^{2} = 931 MeV / u$ .)

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(208)

${}^{235}U \to {}^{140}C e + {}^{94}Z r + n$

$Disintegration energy$

$Q = (m_{R} - m_{P}) c^{2}, m_{R} = 235.0439 u$

$m_{P} = 139.9054 u + 93.9063 u + 1.0086 u = 234.8203 u$

$Q = (235.0439 u - 234.8203 u) c^{2}$

$= 0.2236 c^{2}$

$= 0.2236 \times 931$

$Q = 208.1716$

Solution

Q.
The disintegration energy $Q$ for the nuclear fission of ${}^{235}U \to {}^{140}C e + {}^{94}Z r + n$ is ___ MeV.

(Given atomic masses of ${}^{235}U : 235.0439 u; {}^{140}C e : 139.9054 u; {}^{94}Z r : 93.9063 u; n : 1.0086 u$ , and value of $c^{2} = 931 MeV / u$ .)

Ans.
(208)

${}^{235}U \to {}^{140}C e + {}^{94}Z r + n$

$Disintegration energy$

$Q = (m_{R} - m_{P}) c^{2}, m_{R} = 235.0439 u$

$m_{P} = 139.9054 u + 93.9063 u + 1.0086 u = 234.8203 u$

$Q = (235.0439 u - 234.8203 u) c^{2}$

$= 0.2236 c^{2}$

$= 0.2236 \times 931$

$Q = 208.1716$

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Solution

Q. The disintegration energy Q for the nuclear fission of U235→C140e+Z94r+n is ___ MeV. (Given atomic masses of U235:235.0439u; C140e:139.9054u; Z94r:93.9063u; n:1.0086u, and value of c2=931 MeV/u.)

Ans. (208) U235→C140e+Z94r+n Disintegration energy Q=(mR-mP)c2, mR=235.0439 u mP=139.9054 u+93.9063 u+1.0086 u=234.8203 u Q=(235.0439 u-234.8203 u)c2 =0.2236 c2 =0.2236×931 Q=208.1716

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Q.
The disintegration energy $Q$ for the nuclear fission of ${}^{235}U \to {}^{140}C e + {}^{94}Z r + n$ is ___ MeV.

(Given atomic masses of ${}^{235}U : 235.0439 u; {}^{140}C e : 139.9054 u; {}^{94}Z r : 93.9063 u; n : 1.0086 u$ , and value of $c^{2} = 931 MeV / u$ .)

Ans.
(208)

${}^{235}U \to {}^{140}C e + {}^{94}Z r + n$

$Disintegration energy$

$Q = (m_{R} - m_{P}) c^{2}, m_{R} = 235.0439 u$

$m_{P} = 139.9054 u + 93.9063 u + 1.0086 u = 234.8203 u$

$Q = (235.0439 u - 234.8203 u) c^{2}$

$= 0.2236 c^{2}$

$= 0.2236 \times 931$

$Q = 208.1716$