a nucleus(mass no.238) initially at rest decay into another nucleus (mass no.234) emitting an alpha particle (mass no.4) with the speed of 1.17×10^7metre per second find the recoil speed of remaining nucleus
Answers
Intially the momentum of mass no 238 i.e. uranium nucleus is zero as it at rest . After emission of the alpha particle the sum of momenta of alpha particle and remaining mass no 234 i.e. thorium is also zero ,because the radioactive emmision is purely internal process and no external force acts upon it . thus
multiplication of mass alpha and velocityalpha+ multiplication of massthorium and velocitythorium =0 (according to law of conservation of momentum) , so
velocity of thorium = -malpha particle / mthorium into velocityalpha
we know that massalpha / mass thorium = 4/234
given valpha = 1.17 into 107 so velocity of thorium = -4/234 into 1.17 into 107
= - 2 into 105 m/s here minus represents direction
Answer:
Explanation:
consevation of linear momentum
a nucleus (mass no 238)initially at rest decay into another nucleus(mass no 234)emitting an alpha particle(mass no 4)with speed of 1.17*10to the power 7m/sec.find the speed of remaining nucleus
7 years ago
Answers : (1)
Intially the momentum of mass no 238 i.e. uranium nucleus is zero as it at rest . After emission of the alpha particle the sum of momenta of alpha particle and remaining mass no 234 i.e. thorium is also zero ,because the radioactive emmision is purely internal process and no external force acts upon it . thus
multiplication of mass alpha and velocityalpha+ multiplication of massthorium and velocitythorium =0 (according to law of conservation of momentum) , so
velocity of thorium = -malpha particle / mthorium into velocityalpha
we know that massalpha / mass thorium = 4/234
given valpha = 1.17 into 107 so velocity of thorium = -4/234 into 1.17 into 107
= - 2 into 105 m/s here minus represents direction