The nucleus 23 10 Ne decays by β – emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that: m ( 23 10 Ne) = 22.994466 u m ( 23 11 Ne) = 22.089770 u.

Question

Philoid · Accepted Answer

For β^- decay,the number of proton increases by 1.The reaction is given as:

Here the electron masses gets cancelled as Na has one more electron than Ne hence,

Q = (22.994466-22.089770)u × c²

= 0.00469 × 931.5 MeV

= 4.37 MeV

The maximum k.E of the emitted electrons are comparable to the Q value; Hence the maximum K.E of the electrons emitted will be = 4.37 MeV

The nucleus 2310Ne decays by β– emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:m (2310Ne) = 22.994466 um (2311Ne) = 22.089770 u.

The nucleus ²³₁₀Ne decays by β^– emission. Write down the β-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
m (²³₁₀Ne) = 22.994466 u

m (²³₁₁Ne) = 22.089770 u.