`._(10)^(23)Ne` का नाभिक `beta^(-)` उत्सर्जन के साथ क्षयित होता है। इस `beta`-क्षय के लिए समीकरण लिखिए और उत्‍सर्जित इलेक्ट्रॉनों की अधिकतम गतिज ऊर्जा ज्ञात कीजिए।
`m(._(10)^(23)Ne)=22.994466u`
`m(._(11)^(23)Na)=22.989770u`

The nucleus ""_(10)^(23)Ne decays by beta^(-) emission. Write down the beta -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 .

The nucleus ""_(10)^(23)Ne decays by beta^(-) – mission. Write down the beta -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.

The nucleus ""_(10)^(23)Ne decays by beta^(-) – mission. Write down the beta -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.

The nucleus ._(10)^(23)Ne decays by beta - emission. Write down the beta - decay. Write down the beta - 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)=22.089770 u .

The nucleus "_10^23Ne decays by beta^- emission. Write down the beta -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.

The nucleus .^(23)Ne deacays by beta -emission into the nucleus .^(23)Na . Write down the beta -decay equation and determine the maximum kinetic energy of the electrons emitted. Given, (m(._(10)^(23)Ne) =22.994466 am u and m (._(11)^(23)Na =22.989770 am u . Ignore the mass of antineuttino (bar(v)) .

The nucleus .^(23)Ne deacays by beta -emission into the nucleus .^(23)Na . Write down the beta -decay equation and determine the maximum kinetic energy of the electrons emitted. Given, (m(._(11)^(23)Ne) =22.994466 am u and m (._(11)^(23)Na =22.989770 am u . Ignore the mass of antineuttino (bar(v)) .