Suppose some nucleus undergoes decay, releasing 5.4 MeV of energy. If the daught

Question

Suppose some nucleus undergoes decay, releasing 5.4 MeV of energy. If the daughter product of this reaction is 93X237 (atomic mass = 237.04816) , it will recoil away as the particle leaves. Determine (a) the energy of the daughter product and (b) the energy of the particle (atomic mass = 4.002603 u). Assume that the energy of each particle is kinetic energy, and ignore any other small amounts of energy that might be carried away by other emissions. In addition, ignore relativistic effects.

Explanation / Answer

let,

m1=237.04816*u=237.04816*1.66*10^-27

m1=3.93*10^-25kg

m2=4.002603*u=4.002603*1.66*10^-27

m2=6.64*10^-27kg

energy released, U=5.4 Mev

a)

by using law of conservation of momentum,

m1*V1+m2*v2=m1*u1+m2*u2

3.93*10^-25*v1-6.64*10^-25*v2=0 ----(1)

by using law of conservation of energy,

U=1/2*m1*v1^2 + 1/2*m2*v2^2

5.4*1.6*10^-13=1/2*3.93*10^-25*v1^2 + 1/2*6.64*10^-27*v2^2 ---(2)

from the above equation (1) and (2)

v1=2.09*10^6 m/sec

and

v2=1.24*10^6 m/sec

now,

K.E of the daughter product,

K.E=1/2*m1*v1^2

=1/2*3.93*10^-25*(2.09*10^6)^2

=8.58*10^-14 J

b)

K.E of the particle

=1/2*m2*v2^2

=1/2*6.64*10^-27*(1.24*10^6)^2

=5.105*10^-15 J

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