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प्रश्न
A nucleus \[\ce{^194_82X}\] emits an alpha particle
- What will be the atomic number of the daughter nucleus Y?
- What will be the number of neutrons in the daughter nucleus Y?
- Write a nuclear reaction showing the emission of this particle.
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उत्तर
(a) \[\ce{^194_82X}\] emits an alpha particle.
We know, an Alpha particle consists of two protons and two neutrons \[\ce{(^4_2\alpha})\]. On emitting an alpha particle the atomic number will reduce by 2 and the mass number will reduce by 4.
Thus, an atomic number of the daughter nucleus Y is = 82 - 2 = 80.
Mass number of the daughter nucleus Y is = 194 - 4 = 190.
(b) Number of protons in daughter nucleus Y is = 80.
Thus, number of neutrons in daughter nucleus Y is = 190 - 80 = 110.
(c) Nuclear reaction showing the emission of this particle is, \[\ce{^194_-82X ->[-(4/2 \alpha)] ^190_80Y}\]
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