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))`.

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

Complete the decay reaction ._10Ne^(23) to?+._-1e^0+? Also, find the maximum KE of electrons emitted during this decay. Given mass of ._10Ne^(23)=22.994465 u . mass of ._11Na^(23)=22.989768u .

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.

A nucleus ""_(10)Ne^(23) undergoes beta - decay and becomes ""_(11)Na^(23) . Calculate the maximum kinetic energy of electrons emitted assuming that the daughter nucleus and antineutrino carry negligible kinetic energy. Given mass of ""_(10)Ne^(23) = 22.994466 u and mass of ""_(11)Na^(23) = 22.989770 m

Gold ._(79)^(198)Au undergoes beta^(-) decay to an excited state of ._(80)^(198)Hg . If the excited state decays by emission of a gamma -photon with energy 0.412 MeV , the maximum kinetic energy of the electron emitted in the decay is (This maximum occurs when the antineutrino has negligble energy. The recoil enregy of the ._(80)^(198)Hg nucleus can be ignored. The masses of the neutral atoms in their ground states are 197.968255 u for ._(79)^(198)Hg ).

The radionuclide ._6C^(11) decays according to ._6C^(11)to ._5B^(11) +e^(+) +v : half life =20.3min. The maximum energy of the emitted positron is 0.960 MeV . Given the mass values m(._6C^(11))=11.011434u, m(._6B^(11))=11.009305u Calculate Q and compare it with maximum energy of positron emitted.

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.

Consider the beta decay ^198 Au rarr ^198 Hg ** + Beta^(-1) + vec v . where ^198 Hg^** represents a mercury nucleus in an excited state at energy 1.088 MeV above the ground state. What can be the maximum kinetic energy of the electron emitted? The atomic mass of ^198 Au is 197.968233 u and that of ^198 Hg is 197.966760 u .