Question

Complete the following decay,
23 Ne10-->? +0e-1 + ?
Also find the maximum kinetic energy of the electron emitted during this decay. Given, mass
of Ne23 = 22.994465 u and mass of Na23 = 22.989768 u.
​

Answer 1

The maximum kinetic energy of the electron emitted during this decay is 4.37 MeV.

Explanation:

Decay equation

23Ne10  → 23Na11  + β¯¯ + μ¯¯      ........(1)

Maximum  energy  of  the  emitted  electrons  will  be  equal  to  Q  value  of  the  reaction.

Let's find the kinetic energy of the reaction.

K.E(max)  = Δm ×931 MeV

      = (− 22.989770 + 22.994466 ) ×931 MeV                                            

      =4.37 MeV

Thus the maximum kinetic energy of the electron emitted during this decay is 4.37 MeV.

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