Home
Class 12
PHYSICS
Find the Q-value and the kinetic energy ...

Find the Q-value and the kinetic energy of the emitted a-particle in the a-decay of (a) `""_(88)^(226)Ra` and (b) `(86)^(220)Rn`.
Given `m (""_(88)^(226)Ra) = 226.02540 u, " " m (""_(86)^(222)Rn) = 222.01750 u`,
`m (""_(86)^(220)Rn) = 220.01137u, " " m(""_(84)^(216) Po) = 216.00186u`.

Text Solution

Verified by Experts

(a) `Q = 4.93 MeV, E_a = 4.85 MeV (b) Q = 6.41 MeV, E_a = 6.29 MeV`
Promotional Banner

Topper's Solved these Questions

  • NUCLEI

    NCERT GUJARATI|Exercise ADDITIONAL EXERCISES|9 Videos
  • NUCLEI

    NCERT GUJARATI|Exercise ADDITIONAL EXERCISES|9 Videos
  • MOVING CHARGES AND MAGNETISM

    NCERT GUJARATI|Exercise ADDITIONAL EXERCISES|17 Videos
  • RAY OPTICS AND OPTICAL INSTRUMENTS

    NCERT GUJARATI|Exercise EXERCISES|40 Videos

Similar Questions

Explore conceptually related problems

Under certain circumstances, a nucleus can decay by emitting a particle more massive than an a-particle. Consider the following decay processes: ""_(88)^(223)Ra to ""(82)^(209)Pb + ""_(6)^(14)C . ""_(88)^(223)Ra to ""_(86)^(219)Rn + ""_(2)^(4)He Calculate the Q-values for these decays and determine that both are energetically allowed.

The nucleus ""_(10)^(23)Ne decays by beta^(-) – mission. 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 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.

The mass of a nucleus ._(Z)^(A)X is less that the sum of the masses of (A-Z) number of neutrons and Z number of protons in the nucleus.The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass M can break into two light nuclei of masses m_(1) and m_(2) only if (m_(1)+m_(2)) lt M . Also two light nuclei of masses m_(3) and m_(4) can undergo complete fusion and form a heavy nucleus of mass M'. only if (m_(3)+m_(4)) gt M' . The masses of some neutral atoms are given in the table below: |{:(._(1)^(1)H ,1.007825u , ._(1)^(2)H,2.014102u,._(1)^(3)H,3.016050u,._(2)^(4)He,4.002603u),(._(3)^(6)Li,6.015123u,._(3)^(7)Li,7.016004u,._(30)^(70)Zn,69.925325u, ._(34)^(82)Se,81.916709u),(._(64)^(152)Gd,151.91980u,._(82)^(206)Pb,205.974455u,._(83)^(209)Bi,208.980388u,._(84)^(210)Po,209.982876u):}| Taking kinetic energy ( in KeV ) of the alpha particle, when the nucleus ._(84)^(210)P_(0) at rest undergoes alpha decay, is:

Two stars bound together by gravity orbit othe because of their mutual attraction. Such a pair of stars is referred to as a binary star system. One type of binary system is that of a black hole and a companion star. The black hole is a star that has cullapsed on itself and is so missive that not even light rays can escape its gravitational pull therefore when describing the relative motion of a black hole and companion star, the motion of the black hole can be assumed negligible compared to that of the companion. The orbit of the companion star is either elliptical with the black hole at one of the foci or circular with the black hole at the centre. The gravitational potential energy is given by U=-GmM//r where G is the universal gravitational constant, m is the mass of the companion star, M is the mass of the black hole, and r is the distance between the centre of the companion star and the centre of the black hole. Since the gravitational force is conservative. The companion star and the centre of the black hole, since the gravitational force is conservative the companion star's total mechanical energy is a constant. Because of the periodic nature of of orbit there is a simple relation between the average kinetic energy ltKgt of the companion star Two special points along the orbit are single out by astronomers. Parigee isthe point at which the companion star is closest to the black hole, and apogee is the point at which is the farthest from the black hole. Q. For circular orbits the potential energy of the companion star is constant throughout the orbit. if the radius of the orbit doubles, what is the new value of the velocity of the companion star?

NCERT GUJARATI-NUCLEI-EXERCISE
  1. Obtain the binding energy (in MeV) of a nitrogen nucleus (""(7)^(14)N)...

    Text Solution

    |

  2. Obtain the binding energy of the nuclei ""(26)^(26)Fe and ""(83)^(209)...

    Text Solution

    |

  3. A given coin has a mass of 3.0 g. Calculate the nuclear energy that wo...

    Text Solution

    |

  4. Write nuclear reaction equations for {:((i) a-"decay of " (88)^(226)...

    Text Solution

    |

  5. A radioactive isotope has a half-life of T years. How long will it tak...

    Text Solution

    |

  6. The normal activity of living carbon-containing matter is found to be ...

    Text Solution

    |

  7. Obtain the amount of ""(27)^(60)Co necessary to provide a radioactive ...

    Text Solution

    |

  8. The half-life of ""(38)^(90)Sr is 28 years. What is the disintegration...

    Text Solution

    |

  9. Obtain approximately the ratio of the nuclear radii of the gold isotop...

    Text Solution

    |

  10. Find the Q-value and the kinetic energy of the emitted a-particle in t...

    Text Solution

    |

  11. The radionuclide ""^(11)C decays according to ""(6)^(11)C to ""(5)^(...

    Text Solution

    |

  12. The nucleus ""(10)^(23)Ne decays by beta^(-)– mission. Write down the ...

    Text Solution

    |

  13. The Q value of a nuclear reaction A + b to C + d is defined by Q =...

    Text Solution

    |

  14. Suppose, we think of fission of a ""(26)^(56)Fe nucleus into two equal...

    Text Solution

    |

  15. The fission properties of ""(94)^(239)Pu are very similar to those of ...

    Text Solution

    |

  16. A 1000 MW fission reactor consumes half of its fuel in 5.00 y. How muc...

    Text Solution

    |

  17. How long can an electric lamp of 100W be kept glowing by fusion of 2.0...

    Text Solution

    |

  18. Calculate the height of the potential barrier for a head on collision ...

    Text Solution

    |

  19. From the relation R = R0A^(1//3), where R0 is a constant and A is the ...

    Text Solution

    |

  20. For the b+ (positron) emission from a nucleus, there is another compet...

    Text Solution

    |