NEET 2026 ALL Q #1936
COMPETENCY BASED
APPLY
4 Marks 2026 NTA-RE-NEET-2026 MCQ SINGLE
Consider the following nuclear reaction ${}^{238}U\longrightarrow{}^{234}Th+{}^{4}He$. Take masses of ${}^{238}U$, ${}^{234}Th$ and ${}^{4}He$ as 238.050 u, 234.043 u and 4.003 u, respectively. The Q value for the reaction, in keV, is: [Given: $1~u=931.5~MeV~c^{-2}$]
(A) 3740
(B) 3726
(C) 3730
(D) 3736
Correct Answer: B

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Detailed Solution

Step 1: Calculate the Mass Defect

The mass defect ($ \Delta m $) is the difference between the mass of the reactant and the sum of the masses of the products. $$ \Delta m = M(^{238}U) - [M(^{234}Th) + M(^{4}He)] $$ $$ \Delta m = 238.050 - [234.043 + 4.003] $$ $$ \Delta m = 238.050 - 238.046 = 0.004 \, u $$

Step 2: Calculate the Q-value in MeV

The Q-value is given by $ Q = \Delta m \times 931.5 \, MeV/u $. $$ Q = 0.004 \times 931.5 = 3.726 \, MeV $$

Step 3: Convert MeV to keV

Since $ 1 \, MeV = 1000 \, keV $: $$ Q = 3.726 \times 1000 = 3726 \, keV $$

Final Answer: 3726

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Pedagogical Audit
Bloom's Analysis: This is an APPLY question because the student must utilize the concept of mass-energy equivalence and perform unit conversions to solve a specific nuclear physics problem.
Knowledge Dimension: PROCEDURAL
Justification: The question requires a step-by-step algorithmic approach involving subtraction of masses followed by multiplication with a conversion factor.
Syllabus Audit: In the context of NEET, this is classified as COMPETENCY. This question tests the student's ability to handle numerical data in nuclear reactions, which is a core competency required for the Physics section of the NEET examination.