The nucleus 2310 Ne decays by β–emission. Write down the β-decay equation and determine the maximum kinetic energy of theelectrons emitted. Given that:m (2310 Ne ) = 22.994466 um (2311 Na ) = 22.089770 u.
Given: The mass of N 10 23 e is 22.994466 u and the mass of N 11 23 a is 22.989770 u .
In β − emissions, the numbers of protons are increased by one and one antineutrino and one electron emitted from the parent nucleus.
The reaction of β − emission is given as,
N 10 23 e → N 11 23 a + e − + v ¯ + Q
The Q-value for the reaction is given as,
Q-value = ( m i − m f ) c 2 = [ m ( N 10 23 e ) − m ( N 11 23 a ) ] c 2 (1)
Where, the sum of initial mass is m i and the sum of final mass is m f .
By substituting the given values in equation (1), we get
Q-value = [ 22.994466 − 22.089770 ] u ( c ) 2 = [ 0.004696 c 2 ] u (2)
We know that,
1 u = 931.5 MeV / c 2 (3)
By substituting the values of equation (3) in equation (2), we get
Q-value = 0.004696 × 931.5 = 4.374 MeV
The daughter nuclei is too heavy that it has negligible kinetic energy. The kinetic energy of antineutrino is almost zero.
The total kinetic energy is equal to the Q-value .
K . E . = 4.374 MeV
Thus, the kinetic energy of emitted electrons is 4.374 MeV .
Given: The mass of
In
The reaction of
The
Where, the sum of initial mass is
By substituting the given values in equation (1), we get
We know that,
By substituting the values of equation (3) in equation (2), we get
The daughter nuclei is too heavy that it has negligible kinetic energy. The kinetic energy of antineutrino is almost zero.
The total kinetic energy is equal to the
Thus, the kinetic energy of emitted electrons is
