In a fission event of ²38}₉2}{U} by fast ...

CBSE Class 12 Physics 2022 Term 2 Delhi Set 1 Solved Paper

Question : 6 of 12

Marks: +1, -0

In a fission event of

^{238}_{92}\mathrm{U}

by fast moving neutrons, no neutrons are emitted and final products, after the beta decay of the primary fragments, are

^{140}_{58}\mathrm{Ce}

and

^{99}_{44}\mathrm{Ru}

. Calculate

Q

for this process. Neglect the masses of electrons/ positrons emitted during the intermediate steps. 3

Given:

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

;

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

m(^{99}_{44}\mathrm{Ru}) = 98.90594\,\mathrm{u} ; m(^{1}_{0}\mathrm{n}) = 1.008665\,\mathrm{u}

Solution:

{}^{238}_{92}\mathrm{U} + {}^{1}_{0}\mathrm{n} \rightarrow {}^{239}_{92}\mathrm{U} \rightarrow {}^{140}_{58}\mathrm{Ce} + {}^{99}_{44}\mathrm{Ru} + 10\,{}^{0}_{-1}\mathrm{e}

\Delta m = (238.05079 + 1.008665 - 139.90543 - 98.90549)\,\mathrm{u}

[neglecting mass of electron]

Or.

\Delta m = 0.24835\,\mathrm{u}

\therefore Q = 0.24835 \times 931 = 231.386085 = 231.386\,\mathrm{MeV}

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