Question

A nucleus \[\ce{^194_82X}\] emits an alpha particle

1. What will be the atomic number of the daughter nucleus Y?
2. What will be the number of neutrons in the daughter nucleus Y?
3. Write a nuclear reaction showing the emission of this particle.

Answer 1

(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, \[\ce{^194_-82X ->[-(4/2 \alpha)] ^190_80Y}\]