The mass number and atomic number of a radioactive element Actinium are 227 and 89 respectively. Calculate the number of α and β particles emitted if the mass number and atomic number of the new element Lead are 207 and 82 respectively.
The Correct Answer and Explanation is :
To determine the number of alpha (α) and beta (β) particles emitted during the radioactive decay of Actinium ((^{227}{89}Ac)) to Lead ((^{207}{82}Pb)), we use the following principles:
Step 1: Understanding Alpha and Beta Decay
- An alpha particle (α) consists of 2 protons and 2 neutrons ((^4_2He)), meaning it decreases the atomic number by 2 and the mass number by 4.
- A beta particle (β) represents an electron ((^0_{-1}e)), meaning it increases the atomic number by 1 without affecting the mass number.
Step 2: Changes in Mass and Atomic Numbers
- Change in Mass Number:
[
227 – 207 = 20
]
Each alpha decay reduces the mass number by 4, so the number of α particles emitted is:
[
\frac{20}{4} = 5
]
Thus, 5 α particles are emitted. - Change in Atomic Number:
[
89 – 82 = 7
]
Each alpha decay decreases the atomic number by 2, so the contribution from 5 α decays is:
[
5 \times 2 = 10
]
The total decrease should be 7, so beta decays must compensate by increasing the atomic number. Each β decay increases the atomic number by 1:
[
10 – 7 = 3
]
Thus, 3 β particles are emitted.
Final Answer:
- Number of α particles emitted = 5
- Number of β particles emitted = 3
This ensures that the final nucleus reaches the mass number of 207 and atomic number of 82, corresponding to Lead ((Pb)).
Now, I’ll generate an image to illustrate this decay process.
Here is the scientific diagram illustrating the radioactive decay of Actinium-227 to Lead-207. The image clearly shows the emission of 5 alpha (α) particles and 3 beta (β) particles, along with intermediate decay stages leading to the final stable Lead-207 nucleus. Let me know if you need any modifications or further explanations!