Question

The nucleus of a certain atom has a mass of 3.8x10^-25 (i.e. 10 raised to the power -ve25) kg and is at rest. The nucleus is radioactive and suddenly ejects from itself a particle of mass 6.6 x 10^-27(i.e. 10 raised to the power -ve27) kg and speed 1.5x 10^7(i.e. 10 raised to the power 7) m/s. Find the recoil speed of the nucleus left behind

Answer 1

Initial momentum=Final momentum
Initial momentum=0
Final momentum=MV+mv=0
or
MV=-mv
V=-mv/M
m=6.67*10raised to the power -27
V=1.5*10raised to the power 7
M=3.8*10raised to the power -25
Taking - common
V=-(6.6*1.5*10'-27*1.5'7)/3.8*10'-25
V=2.6*10'-5

Answer 2

Answer:

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