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Question
The nucleus `""_10^23"Ne"` decays by `beta^(-)`emission. Write down the β decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
`"m"(""_10^23 "Ne")` = 22.994466 u
`"m"(""_11^23 "Na")` = 22.989770 u.
Solution
In `beta^(-)` emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus.
`beta^(-)` emission of the nucleus `""_10^23"Ne"` is given as:
\[\ce{^23_10 Ne -> ^23_11Na + e- + \bar{v} + Q}\]
It is given that:
Atomic mass of ,`"m"(""_10^23 "Ne")`= 22.994466 u
Atomic mass of `"m"(""_11^23 "Na")` = 22.989770 u
Mass of an electron, me = 0.000548 u
Q-value of the given reaction is given as:
`"Q" = ["m"(""_10^23"Ne") - ["m"(""_11^23 "Na") + "m"_"e")]]"c"^2`
There are 10 electrons in `""_10^23"Ne"` and 11 electrons in . Hence, the mass of the electron is cancelled in the Q-value equation.
`therefore "Q" = [22.994466 - 22.989770]"c"^2`
`= (0.004696 "c"^2)"u"`
But 1 u = 931.5 MeV/c2
`therefore "Q" = 0.004696 xx 931.5 = 4.374 "MeV"`
The daughter nucleus is too heavy as compared to `"e"^(-)` and `bar"v"`. Hence, it carries negligible energy. The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the emitted electrons is almost equal to the Q-value, i.e., 4.374 MeV.
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