Home
Class 12
PHYSICS
The nucleus .(10)^(23)Ne decays by beta ...

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`.

Text Solution

Verified by Experts

`""_(10)^(23)Ne to ""_(11)^(23)Na + e^(-) + bar(v) + Q , Q = [m_N (""_(10)^(23)Ne) - m_N (""_(11)^(23)Na) - m_(e)] c^2`, where the masses used are masses of nuclei and not of atoms as in Exercise . Using atomic masses `Q = [m (""_(10)^(23)Ne) - m(""_(11)^(23)Na)]c^2`. Note `m_e` has been cancelled. Using given masses, `Q = 4.37 MeV`. As in Exercise , maximum kinetic energy of the electron (max `E_e) = Q = 4.37 MeV`.
Promotional Banner

Similar Questions

Explore conceptually related problems

Obtain the maximum kinetic energy of b-particles, and the radiation frequencies of g decays in the decay scheme shown in Fig. You are given that m(""_(8)^(19)Au) = 197.968233 u m(""_(8)^(19)Hg) =197.966760 u .

The radionuclide ""^(11)C decays according to ""_(6)^(11)C to ""_(5)^(11) B + e^(+) + v, T_(1//2) = 20.3 min The maximum energy of the emitted positron is 0.960 MeV. Given the mass values: m ( ""_(6)^(11)C) = 11.011434 u and m (""_(6)^(11)B ) = 11.009305 u, calculate Q and compare it with the maximum energy of the positron emitted.

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.

Consider the fission of ._(92)^(238)U by fast neutrons. In one fission event, no neutrons are emitted and the final end products, after the beta decay of the primay fragments, are ._(58)^(140)Ce and ._(44)^(99)Ru . Calculate Q for this fission process. The relevant atomic and particle masses are m(._(92)^(238)U)=238.05079 u m(._(58)^(140)Ce)=139.90543 u m(._(44)^(99)Ru)=98.90594 u

A nuclear af rest undergoes a decay emitting and alpha- particle of de-Broglie wavelength lamda=5.76xx10^(-15) meter. he mass of the daughter nucleus is 223.610 amu and that of the alpha - particle is 4.002 a.m.u. (1"amu"=931.470(MeV)/(C^(2))) Momentum of the alpha - particle is nearly