Home
Class 12
PHYSICS
Consider the fission of " "(92)^(238)U b...

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 primary 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`.

Text Solution

Verified by Experts

For the fission `" "_(92)^(238)U +n to " "_(58)^(140)Ce + " "_(44)^(99)Ru +Q`
`Q = [m (" "_(92)^(238)U) + m_(n) -m (" "_(58)^(140)Ce) - m(" "_(44)^(99)Ru)]c^(2)= [238.05079 + 1.00867 - 139.90543 - 98.90594] u xx c^(2) = 0.24809 u xx 931.5 MeV = 231.1 MeV`.
Promotional Banner

Topper's Solved these Questions

  • NUCLEI

    U-LIKE SERIES|Exercise CBSE Based/Source -Based Integrated Questions|3 Videos

  • NUCLEI

    U-LIKE SERIES|Exercise Multiple Choice Questions|30 Videos

  • NUCLEI

    U-LIKE SERIES|Exercise Self Assessment Test|10 Videos

  • MOVING CHARGES AND MAGNETISM

    U-LIKE SERIES|Exercise SELF ASSESSMENT TEST (SECTION D)|1 Videos

  • RAY OPTICS AND OPTICAL INSTRUMENTS

    U-LIKE SERIES|Exercise SELF ASSESSMENT TEST (SECTION B)|3 Videos

Similar Questions

Explore conceptually related problems

Consider the fission ._(92)U^(238) by fast neutrons. In one fission event, no neutrons are emitted and the final stable and products, after the beta decay of the primary fragments are ._(58)Ce^(140) and ._(44)Ru^(99) . Calculate Q for this fission process, The relevant atomic and particle masses are: m(._(92)U^(238))=238.05079u, m(._(58)Ce^(140))=139.90543 u, m(._(34)Ru^(99))=98.90594 u

Calculate the number of neutrons in the remaining atoms after the emission of an alpha particle from ._(92)U^(238) atom.

The atomic number (A) and mass number (M) of the nuclide formed when three alpha (alpha) and 2 beta (beta) particles are emitted from ""_(92)^(238)U is

Consider one of the fission reactions of U^(235) by thernmal neutrons : ._(92)U^(235) + n rarr ._(38)Sr^(94) + ._(54)Ce^(140) + 2n The fission fragments are, however, not stable. They unaergo successive beta- decays unit ._(38)Sr^(94) becomes ._(40)Zr^(94) and ._(54)Xe^(140) becomes ._(58)Ce^(140) . Estimate the total energy released in the process. Is all that energy available as kinetic energy of the fission products (Zr and Ce)? You are given that m (U^(235)) = 255.0439 am u," " m_(n) = 1.00866 am u " " m(Zr^(94)) = 93.9065 am u, " " m(Ce^(140)) = 139.9055 am u

The atomic number (A) and mass number (M) of the nuclide formed where three alpha (alpha) and two (beta) particles are emitted from ._(92)^(238)U

Consider the case of bombardment of U^(235) nucleus with a thermal neutron. The fission products are Mo^(95) & La^(139) and two neutrons. Calculate the energy released by one U^(235) nucleus. (Rest masses of the nuclides are U^(235)=235.0439 u, ._(0)^(1)n=1.0087u, Mo^(95)=94.9058 u, La^(139).9061u) .

After the emission of one alpha - particle followed by two beta -particles from "" _(92) ^(238) U, the number of neutrons in the newly formed nucleus is

A uranium nucleus ._(92)U^(238) emits and alpha -particle and a beta -particle in succession. The atomic number and mass number of the final nucleus will be