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The nuclear reaction accompanied with emission of neutron(s) are
(i) 13Al27 + 2He4 → 15P30
(ii) 6C12 + 1H1 → 7N13
(iii) 15P30 → 14Si30 + 1e0
(iv) 96Am241 + 2He4 → 97Bk244 + 1e0
(i) 13Al27 + 2He4 → 15P30
(ii) 6C12 + 1H1 → 7N13
(iii) 15P30 → 14Si30 + 1e0
(iv) 96Am241 + 2He4 → 97Bk244 + 1e0
- a)(i) only
- b)(i) and (iv)
- c)(ii) and (iv)
- d)(iii) only
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
The nuclear reaction accompanied with emission of neutron(s) are(i)13A...
Explanation:
Given nuclear reactions:
(i) 13Al27 + 2He4 → 15P30
(ii) 6C12 + 1H1 → 7N13
(iii) 15P30 → 14Si30 + 1e0
(iv) 96Am241 + 2He4 → 97Bk244 + 1e0
Accompanied with emission of neutron(s):
- The reactions (i) and (iv) involve the emission of a neutron.
- In reaction (i), the fusion of 13Al27 and 2He4 results in the formation of 15P30 along with the emission of a neutron.
- In reaction (iv), the fusion of 96Am241 and 2He4 leads to the formation of 97Bk244 along with the emission of a neutron.
Correct answer: Option B (i) and (iv)
Therefore, the correct answer is option B as both reactions (i) and (iv) are accompanied by the emission of neutron(s).
Given nuclear reactions:
(i) 13Al27 + 2He4 → 15P30
(ii) 6C12 + 1H1 → 7N13
(iii) 15P30 → 14Si30 + 1e0
(iv) 96Am241 + 2He4 → 97Bk244 + 1e0
Accompanied with emission of neutron(s):
- The reactions (i) and (iv) involve the emission of a neutron.
- In reaction (i), the fusion of 13Al27 and 2He4 results in the formation of 15P30 along with the emission of a neutron.
- In reaction (iv), the fusion of 96Am241 and 2He4 leads to the formation of 97Bk244 along with the emission of a neutron.
Correct answer: Option B (i) and (iv)
Therefore, the correct answer is option B as both reactions (i) and (iv) are accompanied by the emission of neutron(s).
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Community Answer
The nuclear reaction accompanied with emission of neutron(s) are(i)13A...
- For the nuclear reaction
(i) 13Al27 + 2He4 → 15P30 + X
let X number of n particles are lost. - The reaction can be represented as
13Al27 + 2He4 → 15P30 + 1 1n0
When 13Al27 changes to 15P30, the mass number and atomic number decrease by 1 and 0 units. - This satisfies the nuclear reaction,
13Al27 + 2He4 → 15P30 + 1 1n0 - Also, for the nuclear reaction,
(iv) 96Am241 + 2He4 → 97Bk244 + 1e0 + Z
A neutron emission satisfies the reaction
96Am241 + 2He4 → 97Bk244 + 1e0 + 1n0
Now, for the nuclear reactions (ii) and (iii) does not emotts any neutrons.
Hence, the nuclear reaction accompanied with emission of neutron(s) are (i) and (iv).
A neutron emission satisfies the reaction
96Am241 + 2He4 → 97Bk244 + 1e0 + 1n0
Now, for the nuclear reactions (ii) and (iii) does not emotts any neutrons.
Hence, the nuclear reaction accompanied with emission of neutron(s) are (i) and (iv).
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The nuclear reaction accompanied with emission of neutron(s) are(i)13Al27+2He4→15P30(ii)6C12+1H1→7N13(iii)15P30→14Si30+1e0(iv)96Am241+2He4→97Bk244+1e0a)(i) onlyb)(i) and (iv)c)(ii) and (iv)d)(iii) onlyCorrect answer is option 'B'. Can you explain this answer?
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