Mass number of a radioactive element is 232 and its atomic number is 90. When this element undergoes certain nuclear reactions, it transforms into an isotope of lead with a mass number 208 and an atomic number 82. Determine the number of alpha and beta decay that can occur.
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Answered by
1
Answer:
ANSWER
90
Th
232
→
82
Pb
208
+n
1
α+n
2
β
−
Mass Balance
(mass of alpha particle =4)
232=208+4n
1
n
1
=6
Now, atomic no./proton balance
90=82+2n
1
−n
2
n
2
=4
Answered by
3
The no of alpha decays is SIX (6) and the no of beta decays is FOUR (4).
- In alpha decay, the mass no decreases by 4 and the atomic no decreases by 2.
- In beta decay, the atomic no increases by 1 and mass no remains the same.
- The difference in mass no = 232 - 208 = 24
⇒ The no of alpha decays = 24/4 = 6
- For 6 alpha decays, the decrease in atomic no should be = 2 × 6 = 12
- But the decrease in atomic no = 90-82 = 8
⇒ The atomic no is 4 greater than the expected.
- It means that there are 4 beta decays.
- So, in total 6 alpha decays and 4 beta decays can occur.
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