\({ }^{240} X\) ન્યુક્લિયસની પ્રતિ ભંજન છૂટી પડતી ઉર્જા \(200\,MeV\) છે. જો \(120\,g\) શુદ્ધ \({ }^{240} X\) ના બધા જ પરમાણુઓનું ભુંજન થાય, તો છૂટી પડતી ઉર્જા \(..........\times 10^{25}\,MeV\) હશે. \(N _{ A }=6 \times 10^{23}\) આપેલ છે.

A \(5\)
B \(4\)
C \(6\)
D \(3\)

Verified Solution

Answer & Solution

Correct Answer

(C) \(6\)

Step-by-step Solution

Detailed explanation

\(\text { No. of mole }=\frac{120}{240}=\frac{1}{2}\) \(\text { No. of molecules }=\frac{1}{2} \times N _{ A }\) \(\text { Energy released }=\frac{1}{2} \times 6 \times 10^{23} \times 200\) \(=6 \times 10^{25}\,MeV\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From JEE Mains

Explore more questions on app