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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 ${ }_{44}^{99} \mathrm{R}$ u. Calculate $\mathrm{Q}$ for this fission process. The relevant atomic and particlemasses are $\mathrm{m}\left(\begin{array}{c}238 \\ 92\end{array}\right)=238.05079 \mathrm{u}$
$\mathrm{m}\left({ }_{58}^{140} \mathrm{Ce}\right)=139.90543 \mathrm{u}$
$\mathrm{m}\left({ }_{-44}^{99} \mathrm{Ru}\right)=98.90594 \mathrm{u}$
$\mathrm{m}\left({ }_{58}^{140} \mathrm{Ce}\right)=139.90543 \mathrm{u}$
$\mathrm{m}\left({ }_{-44}^{99} \mathrm{Ru}\right)=98.90594 \mathrm{u}$
Solution:
1203 Upvotes
Verified Answer
The reaction is
$$
{ }_{92}^{238} \mathrm{U}+{ }_0^1 \mathrm{n} \longrightarrow{ }_{58}^{140} \mathrm{Ce}+{ }_{44}^{99} \mathrm{Ru}
$$
Mass defect $\Delta \mathrm{m}$
$$
=m\left({ }_{92}^{238} \mathrm{U}\right)+\mathrm{m}\left({ }_0^1 \mathrm{n}\right)-\left[\mathrm{m}\left({ }_{58}^{140} \mathrm{Ce}\right)+\mathrm{m}\left({ }_{44}^{99} \mathrm{Ru}\right)\right]
$$
$=238.05079-[139.90543+98.90594]+1.008665=0.248085$
Energy released $\mathrm{Q}=\Delta \mathrm{m} \mathrm{c}^2=0.248085 \times 931=231.1 \mathrm{MeV}$
$$
{ }_{92}^{238} \mathrm{U}+{ }_0^1 \mathrm{n} \longrightarrow{ }_{58}^{140} \mathrm{Ce}+{ }_{44}^{99} \mathrm{Ru}
$$
Mass defect $\Delta \mathrm{m}$
$$
=m\left({ }_{92}^{238} \mathrm{U}\right)+\mathrm{m}\left({ }_0^1 \mathrm{n}\right)-\left[\mathrm{m}\left({ }_{58}^{140} \mathrm{Ce}\right)+\mathrm{m}\left({ }_{44}^{99} \mathrm{Ru}\right)\right]
$$
$=238.05079-[139.90543+98.90594]+1.008665=0.248085$
Energy released $\mathrm{Q}=\Delta \mathrm{m} \mathrm{c}^2=0.248085 \times 931=231.1 \mathrm{MeV}$
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