Complete the following nuclear reactions: B 10 5 + n 1 0 → He 4 2 + . . . . . . .

The given reaction is B 10 5 + n 1 0 → He 4 2 + . . . . . . . The nuclear reactions obey the laws of conservation of mass and charge. By balancing these two quantities on both sides of the equation, we have, B 10 5 + n 1 0 → He 4 2 + Li 7 3 Here boron is bombarded by the neutrons.

Complete the following nuclear reactions: N 14 7 + He 4 2 → H 1 1 + . . . . . . . .

The given reaction is N 14 7 + He 4 2 → H 1 1 + . . . . . . . . The nuclear reactions obey the laws of conservation of mass and charge. By balancing these two quantities on both sides of the equation, we have, N 14 7 + He 4 2 → H 1 1 + O 17 8 The products are hydrogen and oxygen.

Complete the following nuclear reactions: H 3 1 → H 3 2 + . . . . . . .

The given reaction is H 3 1 → H 3 2 + . . . . . . . The nuclear reactions obey the laws of conservation of mass and charge. By balancing these two quantities on both sides of the equation, we have, H 3 1 → H 3 2 + β 0 - 1 + v ¯ The products are helium and antineutrino.

Complete the following nuclear reactions: Hg 200 80 + H 1 1 → Au 197 79 + . . . . . . . .

The given reaction is Hg 200 80 + H 1 1 → Au 197 79 + . . . . . . . . The nuclear reactions obey the laws of conservation of mass and charge. By balancing these two quantities on both sides of the equation, we have, Hg 200 80 + H 1 1 → Au 197 79 + He 4 2

Complete the following nuclear reactions: Li 7 3 + He 4 2 → B 10 5 + . . . . . . .

The given reaction is Li 7 3 + He 4 2 → B 10 5 + . . . . . . . The nuclear reactions obey the laws of conservation of mass and charge. By balancing these two quantities on both sides of the equation, we have, Li 7 3 + He 4 2 → B 10 5 + n 1 0 It is an example of the emission of neutron.

Complete the following nuclear reactions: C 14 6 → β 0 - 1 + . . . . . .

The given reaction is C 14 6 → β 0 - 1 + . . . . . . The nuclear reactions obey the laws of conservation of mass and charge. By balancing these two quantities on both sides of the equation, we have, C 14 6 → β 0 - 1 + N 14 7 + v ¯ In this reaction a positron and antineutrino is formed.

Complete the following nuclear reactions: n 1 0 + C 12 6 → B 11 5 + . . . . . .

The given reaction is n 1 0 + C 12 6 → B 11 5 + . . . . . . The nuclear reactions obey the laws of conservation of mass and charge. By balancing these two quantities on both sides of the equation, we have, n 1 0 + C 12 6 → B 11 5 + H 2 1

Write the nuclear reaction for β - decay of P 32 15 and β + decay of C 11 6 .

As we know that beta particles are energetic electrons, so in β - decay, the charge number is increased by 1 without any change in the mass number, therefore the reaction can be represented as : P 32 15 → S 32 16 + β 0 - 1 + v ¯ In this reaction, an antineutrino is emitted. In the case of β + (positron) emission, the charge number is increased by 1 , without any change in the mass number. The reaction can be written as : C 11 6 → B 11 5 + β 0 + 1 + v , where ν = neutrino.

E M B I B E

ISC Physics Class 12, Part-2>Atoms and Nuclei>Chapter 14 - Nuclear Structure>QUESTIONS

Kumar Mittal Solutions for Exercise 1: QUESTIONS

Author:Kumar Mittal

Kumar Mittal Physics Solutions for Exercise - Kumar Mittal Solutions for Exercise 1: QUESTIONS

Attempt the free practice questions from Exercise 1: QUESTIONS with hints and solutions to strengthen your understanding. ISC Physics Class 12, Part-2 solutions are prepared by Experienced Embibe Experts.