Question

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Answer 1

Answer:

Option(D)

Explanation:

The kinetic energy (K) is given by

K = p²/2m where p is the momentum and m is the mass.

We have p²/(2×4 amu) = 6.7 MeV, since the mass of the α-particle is 4 amu.

In the case of the daughter nucleus (of mass 218 – 4 = 214 amu),

we have ,

p²/(2×214 amu) = K

From the above two equations we obtain K = 0.125 MeV.

Therefore,

The recoil energy of the daughter nucleus = 0.125 MeV

Hope it helps!

Answer 2

Answer:

hey

D is the answer

No doubt