AIIMS2004Physics-Nuclear Physics

AIIMS 2004 Physics Radioactive Decay MCQ Question

Type: MCQ-numerical-Hard-Class 12

A nucleus of mass number A, originally at rest, emits an α-particle with speed v. The daughter nucleus recoils with a speed

A

2v / (A + 4)

B

4v / (A + 4)

C

v / (A - 4)

D

2v / (A - 4)

Correct Answer

Option C

Detailed Explanation

To solve the problem of a nucleus of mass number AA emitting an α-particle with speed vv, we need to apply the principle of conservation of momentum. Let's break down the problem step-by-step.

Step 1: Understand the system

  1. Initial state: The parent nucleus (mass number AA) is at rest, so its initial momentum is: pinitial=0p_{\text{initial}} = 0

  2. Final state: After emitting an α-particle (which has a mass number of 4), we have two particles:

    • The α-particle with mass number 4 and speed vv.
    • The daughter nucleus with mass number A4A - 4 and an unknown speed VV.

Step 2: Apply conservation of momentum

According to the law of conservation of momentum, the total momentum before the decay must equal the total momentum after the decay. Therefore, we can write:

pinitial=pα+pdaughterp_{\text{initial}} = p_{\text{α}} + p_{\text{daughter}}

This can be expressed mathematically as: 0=(4)(v)+(A4)(V)0 = (4)(v) + (A - 4)(-V)

Here, we take the direction of the α-particle's motion as positive, and thus the daughter's recoil velocity VV is negative.

Step 3: Rearranging the equation

Rearranging the above equation gives: 0=4v(A4)V0 = 4v - (A - 4)V (A4)V=4v(A - 4)V = 4v V=4vA4V = \frac{4v}{A - 4}

Step 4: Evaluate the answer

The speed of the daughter nucleus VV is given by: V=4vA4V = \frac{4v}{A - 4}

This matches option C, confirming that the correct answer is indeed V=4vA4V = \frac{4v}{A - 4}.

Step 5: Clarification of other options

  • Option A: 2vA+4\frac{2v}{A + 4}: This expression does not correctly apply the conservation of momentum principles, as it suggests a relationship that doesn't account for the correct mass ratios.

  • Option B: 4vA+4\frac{4v}{A + 4}: Similar to option A, this does not respect the conservation of momentum since it incorrectly adds the mass numbers, leading to an incorrect denominator.

  • Option D: 2vA4\frac{2v}{A - 4}: This is also incorrect, as it fails to account for the mass of the α-particle properly and reduces the numerator incorrectly.

Conclusion

The application of conservation of momentum has shown that the speed of the daughter nucleus after the emission of the α-particle is 4vA4\frac{4v}{A - 4}. Thus, the correct option is C. This problem beautifully illustrates the principle of conservation of momentum in nuclear decay processes.

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