Ncert Exercises & Solutions

13.27 Consider the fission of ${ }_{92}^{238} \mathrm{U}$ by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primary fragments, are ${ }_{58}^{140} \mathrm{Ce}$ and ${ }_{44}^{99} \mathrm{Ru}$ . Calculate Q for this fission process. The relevant atomic and particle masses are

m( ${ }_{92}^{238} \mathrm{U}$ ) =238.05079 u

m( ${ }_{58}^{140} \mathrm{Ce}$ ) =139.90543 u

m( ${ }_{44}^{99} \mathrm{Ru}$ ) = 98.90594 u

Hint: The Q-value is given by

$E=\Delta mc^2$ .
Step 1: Write the fission reaction for

${ }_{92}^{238} \mathrm{U}$ .
In the fission of

${}_{92}^{238}U, 10 β^{-}$ particles decay from the parent nucles. The nuclear reaction can be written as:

${}_{92}^{238}U + {}_{0}^{1}n \to {}_{58}^{140}C e + {}_{44}^{99}R u + 10 {}_{- 1}^{0}e$

Step 2: Find the Q-value for the fission reaction.
It is given that:

Mass of

${}_{92}^{238}U n u c l e u s, m_{1} = 238.05079 u$

$M a s s o f a {}_{58}^{140}C e n u c l e u s, m_{2} = 139.90543 u$

$M a s s o f a {}_{44}^{99}R u n u c l e u s, m_{3} = 98.90594 u$

$M a s s o f a n e u t r o n {}_{0}^{1}n, m_{4} = 1.008665 u$

$Q = [m' ({}_{92}^{238}U) + m ({}_{0}^{1}n) - m' ({}_{58}^{140}C e) - m' ({}_{44}^{99}R u) - 10 m_{e}] c^{2}$

where, m'=Represents the corresponding atomic masses of the nuclei.

$m' = ({}_{92}^{238}U) = m_{1} - 92 m e, m' ({}_{58}^{140}C e) = m_{2 - 58} m_{e'} m' ({}_{44}^{99}R u) = m_{3} - 44 m_{e} a n d m' ({}_{0}^{1}n) = m_{4}$

$Q = [m_{1} - 92 m_{e} + m_{4} - m_{2} + 58 m_{e} - m_{3} + 44 m_{e} - 10 m_{e}] c^{2}$

$= [m_{1} + m_{4} - m_{2} - m_{3}] c^{2}$

$= [238.0507 + 1.008665 - 139.90543 - 98.90594] c^{2}$

$= [0.247995 c^{2}] u$

$B u t 1 u = 931.5 M e V / c^{2}$

$∴ Q = 0.247995 \times 931.5 = 231.007 M e V$

$H e n c e, t h e Q - v a l u e o f t h e f i s s i o n p r o c e s s i s 231.007 M e V .$

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