Question

PLEASE ANSWER THIS FAST........
A nucleus of a certain atom has a mass of 3.8*10^-25 kg and is at rest. The nucleus is radioactive and suddenly ejects from itself a particle of mass 6.6*10^-27 kg and speed 1.5*10^7m/s. Find the recoil speed of the nucleus left behind.

Answer 1

Heres the answer..............

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Answer 2

Answer: The recoil speed of the nucleus = $v =- 2.7 \times 10 ^5 m/s.$

Given data:

The mass of nucleus of a certain atom = $3.8 x 10 ^{-25} kg$  Radioactive eject a mass of $6.6 x 10 ^{-27} kg$ Speed = $1.5 x 10 ^7 m/s.$

The recoil velocity = ?

The remaining mass = $3.8 x 10 ^{-25} - 6.6 x 10 ^{-27} = 3.734 x 10 ^{-25}$

Therefore, Pt = 0 initially, after radioactivity

=> $Pt = ( 6.6 x 10 ^{-27} ) \times (1.5 \times 10 ^7 ) + (3. 734 \times 10 ^{-25}) \times v$

=> $0 = 9.9 \times 10 ^{-20} + 3.734 \times 10 ^{-25} v$

=> $v =- 2.7 \times 10 ^5 m/s.$