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A nucleus .ZX^A emits 9 alpha-particles ...

A nucleus `._ZX^A` emits `9 alpha-`particles and `5p` particle. The ratio of total protons and neutrons in the final nucleus is.

A

`((Z-13))/((A-36))`

B

`((Z-13))/((A-Z-13))`

C

`((Z-13))/((A-Z-23))`

D

`((Z-13))/((A-Z-23))`

Text Solution

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The correct Answer is:
To solve the problem step by step, we will analyze the emission of alpha and beta particles from the nucleus \( _Z^A \) and calculate the final number of protons and neutrons. ### Step 1: Understand the emissions 1. **Alpha particles**: Each alpha particle (\( \alpha \)) consists of 2 protons and 2 neutrons, and has a mass number of 4. If 9 alpha particles are emitted, the total change in protons and mass number is: - Protons lost: \( 9 \times 2 = 18 \) - Mass number lost: \( 9 \times 4 = 36 \) ### Step 2: Calculate the new values after alpha emission 2. **New atomic number (Z)** after emitting 9 alpha particles: \[ Z' = Z - 18 \] 3. **New mass number (A)** after emitting 9 alpha particles: \[ A' = A - 36 \] ### Step 3: Understand beta particle emissions 4. **Beta particles**: Each beta particle (\( \beta \)) is an electron with a charge of -1 and mass of 0. Emitting a beta particle increases the atomic number by 1. If 5 beta particles are emitted: - Protons gained: \( 5 \) ### Step 4: Calculate the new values after beta emission 5. **New atomic number (Z)** after emitting 5 beta particles: \[ Z'' = (Z - 18) + 5 = Z - 13 \] 6. **Mass number (A)** remains unchanged after beta emissions: \[ A'' = A - 36 \] ### Step 5: Calculate the number of neutrons 7. **Number of neutrons (N)** in the final nucleus can be calculated using the formula: \[ N = A'' - Z'' \] Plugging in the values we found: \[ N = (A - 36) - (Z - 13) = A - 36 - Z + 13 = A - Z - 23 \] ### Step 6: Calculate the ratio of protons to neutrons 8. **Protons (P)** in the final nucleus: \[ P = Z - 13 \] 9. **Ratio of protons to neutrons**: \[ \text{Ratio} = \frac{P}{N} = \frac{Z - 13}{A - Z - 23} \] ### Final Result The ratio of total protons to neutrons in the final nucleus is: \[ \text{Ratio} = \frac{Z - 13}{A - Z - 23} \]

To solve the problem step by step, we will analyze the emission of alpha and beta particles from the nucleus \( _Z^A \) and calculate the final number of protons and neutrons. ### Step 1: Understand the emissions 1. **Alpha particles**: Each alpha particle (\( \alpha \)) consists of 2 protons and 2 neutrons, and has a mass number of 4. If 9 alpha particles are emitted, the total change in protons and mass number is: - Protons lost: \( 9 \times 2 = 18 \) - Mass number lost: \( 9 \times 4 = 36 \) ### Step 2: Calculate the new values after alpha emission ...
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