# The Nucleus _10^23ne Decays by β−β-emission. Write Down the β Decay Equation and Determine the Maximum Kinetic Energy of the Electrons Emitted. Given that - Physics

Numerical

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, m= 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.

Concept: Atomic Masses and Composition of Nucleus
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#### APPEARS IN

NCERT Physics Part 1 and 2 Class 12
Chapter 13 Nuclei
Exercise | Q 13.14 | Page 463
NCERT Class 12 Physics Textbook
Chapter 13 Nuclei
Exercise | Q 14 | Page 463

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