Neon-23 decays in the following way,
`_10^23Nerarr_11^23Na+ _(-1)^0e+barv`
Find the minimum and maximum kinetic energy that the beta particle`(_(-1)^0e)`can have. The atomic masses of `^23Ne` and `^23 Na` are `22.9945u` and `22.9898u`, respectively.

Neon -23 decays in the following way ""_(10)^(23)Nerarr""_(11)^(23)Ne+""_(-1)e^0+barv. Find the minimum and maximum kinetic energy that the beta particle (""_(-1)e^(0)) can have . The atomic masses of ""^(23)Ne and ""^(23)Na are 22.9945 u and 22.9898 u, respectively.

In beta decay, ""_(10)Ne^(23) is converted into ""_(11)Na^(23) . Calculate the maximum kinetic energy of the electrons emitted, given atomic masses of ""_(10)Ne^(23) and ""_(11)Na^(23) arc 22.994466 u and 22.989770 u, respectively.

.^(23)Ne decays to .^(23)Na by negative beta emission. What is the maximum kinetic energy of the emitter electron ?

Calculate the maximum kinetic energy of the beta particle (positron) emitted in the decay of "_11 ^ 22 Na , given the mass of "_11 ^22 Na -21.994437u "_10 ^22 Ne=21.991385u and m_e=0.00055u

.^(23)Ne decays to .^(23)Na by negative beta emission. What is the maximum kinetic enerfy of the emitter electron ?

.^(23)Ne decays to .^(23)Na by negative beta emission. What is the maximum kinetic enerfy of the emitter electron ?

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)) .