$\text { Match the LIST-I with LIST-II }$
| List - I |
List - II |
||
|---|---|---|---|
| A. | $$ { }_0^1 \mathrm{n}+{ }_{92}^{235} \mathrm{U} \rightarrow{ }_{54}^{140} \mathrm{Xe}+{ }_{38}^{94} \mathrm{Sr}+2{ }_0^1 \mathrm{n} $$ |
I. | $ \text { Chemical reaction } $ |
| B. | $ 2 \mathrm{H}_2+\mathrm{O}_2 \rightarrow 2 \mathrm{H}_2 \mathrm{O} $ |
II. | $ \text { Fusion with +ve } \mathrm{Q} \text { value } $ |
| C. | $$ { }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H} \rightarrow{ }_2^3 \mathrm{He}+{ }_0^1 \mathrm{n} $$ |
III. | $ \text { Fission } $ |
| D. | $$ { }_1^1 \mathrm{H}+{ }_1^3 \mathrm{H} \rightarrow{ }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H} $$ |
IV. | $ \text { Fusion with -ve } Q \text { value } $ |
Solution
<p>Here’s the correct matching:</p>
<p><p>A. </p>
<p>$$_0^1\text{n}+_{92}^{235}\text{U}\to_{54}^{140}\text{Xe}+_{38}^{94}\text{Sr}+2\,_0^1\text{n}$$ </p>
<p>–– This is nuclear fission ⇒ III.</p></p>
<p><p>B. </p>
<p>$2\text{H}_2+\text{O}_2\to2\text{H}_2\text{O}$ </p>
<p>–– This is a chemical reaction ⇒ I.</p></p>
<p><p>C. </p>
<p>$_1^2\text{H}+_1^2\text{H}\to_2^3\text{He}+_0^1\text{n}$ </p>
<p>–– This D–D fusion releases about 3.3 MeV ⇒ fusion with +ve Q ⇒ II.</p></p>
<p><p>D. </p>
<p>$_1^1\text{H}+_1^3\text{H}\to_1^2\text{H}+_1^2\text{H}$ </p>
<p>–– The mass of products exceeds reactants ⇒ requires energy ⇒ fusion with –ve Q ⇒ IV.</p></p>
<p>So the answer is </p>
<p>Option B: A–III, B–I, C–II, D–IV.</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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