Home
Class 12
PHYSICS
Neon-23 decays in the following way, ....

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.

Text Solution

Verified by Experts

here ,atomic masses are given (not the nuclear masses ),But still we can electrons get cancelled both sides thus ,
Mass defect , `Delta m-m(""_(10)Ne^(23))-M(""_(11)Na^(23))`
`=(22.995-22.9898)=0.0047u`
Kinetic energy is given by `Q=Delta m.c^(2)`
`therefore Q-(0.0047 u )(931.5 Me V//u)=4.4 MeV `
hence , the energy of beta particles can range form O to 4.4 MeV .
Promotional Banner

Topper's Solved these Questions

  • NUCLEI

    DC PANDEY ENGLISH|Exercise checkpoint 13.1|10 Videos

  • NUCLEI

    DC PANDEY ENGLISH|Exercise checkpoint 13.2|15 Videos

  • MODERN PHYSICS - 2

    DC PANDEY ENGLISH|Exercise Level 2 Subjective|10 Videos

  • RAY OPTICS

    DC PANDEY ENGLISH|Exercise Medical entrance gallary|76 Videos

Similar Questions

Explore conceptually related problems

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

Find the maximum energy that a beta particle can have in the following decay ^176 Lu rarr ^176 Hf + e + vec v . Alomic mass of ^176 Lu is 175.942694 u and that of ^176 Hf is 175.941420 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)) .

Consider the beta decay of an unstable ._(6)^(14)C nuleus initially at rest: ._(6)^(14)C rarr ._(7)^(14)N +._(-1)^(0)e +v._(e) . Is it possible for the maximum kinetic energy of the emiited beta particle to be exactly equal to Q ?

._^(239)Pu._(94) is undergoing alpha-decay according to the equation ._(94)^(239)Pu rarr (._(97)^(235)U) +._2^4 He . The energy released in the process is mostly kinetic energy of the alpha -particle. However, a part of the energy is released as gamma rays. What is the speed of the emiited alpha -particle if the gamma rays radiated out have energy of 0.90 MeV ? Given: Mass of ._(94)^(239)Pu =239.05122 u , mass of (._(97)^(235)U)=235.04299 u and mass of ._1^4He =4.002602 u (1u =931 MeV) .

Complete the decay reaction ._10Na^(23) to?+._-1e^0+? Also, find the maximum KE of electrons emitted during this decay. Given mass of ._10Na^(23)=22.994465 u . mass of ._11Na^(23)=22.989768u .

The three stable isotopes of neon ._10Ne^(20), ._10Ne^(21) and ._10Ne^(22) have respective abundances of 90.51%,0.27% and 9.22% . The atomic masses of the three isotopes are 19.99u, 20.99u and 21.99u respectively. Obtain the average atomic mass of neon.

The isotope (5)^(12) B having a mass 12.014 u undergoes beta - decay to _(6)^(12) C _(6)^(12) C has an excited state of the nucleus ( _(6)^(12) C ^(**) at 4.041 MeV above its ground state if _(5)^(12)E decay to _(6)^(12) C ^(**) , the maximum kinetic energy of the beta - particle in unit of MeV is (1 u = 931.5MeV//c^(2) where c is the speed of light in vaccuum) .

Find the equation of the ellipse in the following case: focus is (-2,3) directrix is x+y=0 and e=1/2

DC PANDEY ENGLISH-NUCLEI-C MADICAL ENTRANCES GALLERY
  1. Neon-23 decays in the following way, .10^23Nerarr11^23Na+ (-1)^0e+ba...

    Text Solution

    |

  2. The half-life of a radioactive substance is 30 minutes, The time (in m...

    Text Solution

    |

  3. The kinetic energy of alpha-particle emitted in the alpha-decay of ""(...

    Text Solution

    |

  4. If radius of the .(13)^(27)Al nucleus is taken to be R(AI), then the r...

    Text Solution

    |

  5. Two spherical nuclei have mass number 216 and 64 with their radii R(1)...

    Text Solution

    |

  6. In a nuclear reactor, the number of U^(235) nuclei undergoing fissions...

    Text Solution

    |

  7. Nuclear fusion is not found in

    Text Solution

    |

  8. Obtain approx. the ratio of the nuclear radii of the gold isotope .79A...

    Text Solution

    |

  9. A radioactive element X disintegrates successively Xoverset(beta^(-)...

    Text Solution

    |

  10. The energy released by the fission of one uranium atom is 200 MeV. The...

    Text Solution

    |

  11. How many alpha and beta^(-) decays does U^(238) experiences before tu...

    Text Solution

    |

  12. 20% of a radioactive substances decay in 10 days . Calculate the amoun...

    Text Solution

    |

  13. A nucleus splits into two nuclear parts having radii in the ratio 1:2 ...

    Text Solution

    |

  14. What is the energy released by fassion of 1 g of U^(235)? (Assume 200 ...

    Text Solution

    |

  15. The half-life of radium is 1620 yr and its atomic weight is 226 kg per...

    Text Solution

    |

  16. The relationship between decay constant lamda and half-life T of a rad...

    Text Solution

    |

  17. The binding energy per nucleon of .(3)^(7) Li and .(2)^(4)He nuclei ar...

    Text Solution

    |

  18. For the stabilty of any nucleus,

    Text Solution

    |

  19. A uranium nucleus (atomic number 92, mass number 231) emits an alpha-p...

    Text Solution

    |

  20. A radio isotope X with a half life 1.4xx10^(9) yr decays of Y which is...

    Text Solution

    |

  21. After 300 days, the activity of a radioactive sample is 5000 dps (disi...

    Text Solution

    |