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which a U^(238) nucleus original at rest...

which a `U^(238)` nucleus original at rest , decay by emitting an alpha particle having a speed `u` , the recoil speed of the residual nucleus is

A

`(4v)/(238)`

B

`-(4v)/(238)`

C

`v/4`

D

`-(4v)/(234)`

Text Solution

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The correct Answer is:
To solve the problem of finding the recoil speed of the residual nucleus after the decay of a Uranium-238 nucleus emitting an alpha particle, we can use the principle of conservation of momentum. Here’s a step-by-step solution: ### Step 1: Understand the Initial Conditions - The Uranium-238 nucleus is initially at rest, which means its initial velocity \( V_i = 0 \). - The mass of the Uranium-238 nucleus is \( m_{U} = 238 \) units (in atomic mass units, amu). - An alpha particle is emitted with a mass \( m_{\alpha} = 4 \) amu and speed \( u \). ### Step 2: Write the Conservation of Momentum Equation According to the conservation of momentum: \[ \text{Initial Momentum} = \text{Final Momentum} \] Since the Uranium nucleus is at rest initially, the initial momentum is: \[ 0 = m_{U} \cdot V_i \] After the decay, the final momentum consists of the momentum of the emitted alpha particle and the recoil momentum of the residual nucleus (Uranium-234): \[ 0 = m_{\alpha} \cdot u + m_{residual} \cdot V \] Where: - \( m_{residual} = 238 - 4 = 234 \) amu (mass of the residual nucleus) - \( V \) is the recoil speed of the residual nucleus. ### Step 3: Substitute Known Values Substituting the known values into the momentum equation: \[ 0 = (4 \cdot u) + (234 \cdot V) \] Rearranging gives: \[ 234 \cdot V = -4 \cdot u \] ### Step 4: Solve for the Recoil Speed \( V \) To find \( V \), we divide both sides by 234: \[ V = -\frac{4u}{234} \] ### Step 5: Simplify the Expression This can be simplified further: \[ V = -\frac{2u}{117} \] ### Conclusion The recoil speed of the residual nucleus after the alpha decay is: \[ V = -\frac{4u}{234} \] The negative sign indicates that the recoil speed is in the opposite direction to the emitted alpha particle. ### Final Answer The recoil speed of the residual nucleus is: \[ V = -\frac{4u}{234} \]
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