The nucleus ${}_{10}^{23}{Ne}$ decays by β^- emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted from the following data:
$m ({}_{10}^{23}{Ne}) = 22.994466 u m ({}_{11}^{23}{Na}) = 22.989770 u$

Solution

The $β - decay$ of ${}_{10}^{23}{Ne}$ may be represented as:

${}_{10}^{23}{Ne} \to {}_{11}^{23}{Na} - {}_{- 1}^{0}e + \bar{v} + Q$

Ignoring the rest mass of antineutrino $(\bar{v})$ and electron , we get

Mass defect,
$∆ m = m ({}_{10}^{23}{Ne}) - m ({}_{11}^{23}{Na}) = 22.994466 - 22.989770 = 0.004696 u$

$∴$ $Q = 0.004696 \times 931 MeV = 4.372 MeV .$

This energy of 4.3792 MeV, is shared by $e^{-} and \bar{v}$ pair because, ${}_{11}^{23}{Na}$ is very massive.

The maximum K.E. of $e^{-} = 4.372 MeV,$ when energy carried by $\bar{v}$ is zero.

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Wave Optics

The nucleus ${}_{10}^{23}{Ne}$ decays by β^- emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted from the following data:
$m ({}_{10}^{23}{Ne}) = 22.994466 u m ({}_{11}^{23}{Na}) = 22.989770 u$

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Wave Optics

The nucleus Ne1023 decays by β- emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted from the following data:m(Ne1023) = 22.994466 um(Na1123) = 22.989770 u

Some More Questions From Wave Optics Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

The nucleus ${}_{10}^{23}{Ne}$ decays by β^- emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted from the following data:
$m ({}_{10}^{23}{Ne}) = 22.994466 u m ({}_{11}^{23}{Na}) = 22.989770 u$