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 Multiple Choice QuestionsMultiple Choice Questions

171.

Fission of nuclei is possible because the binding energy per nucleon in them

  • increases with mass number at high mass numbers

  • decreases with mass number at high mass numbers

  • increases with mass number at low mass numbers 

  • decreases with mass number at low mass numbers


172.

In any fission process the ratio Mass of parent nucleusMass of fission products is

  • less than 1

  • greater than 1

  • greater than 1

  • depends on the mass of parent nucleus


173.

In radioactive decay process, the negatively charged emitted -particles are 

  • the electrons present inside the nucleus

  • the electrons produced as a result of the decay of neutrons inside the nucleus

  • the electrons produced as a result of collisions between atoms

  • the electrons orbiting around the nucleus


174.

Two radioactive substances A and B have decay constants 5 λ and λ respectively. At t = 0 they have the same number of nuclei. The ratio of number of nuclei of A to those of B will be 1e2 after a time interval

  • 14λ

  • 12λ


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175.

If Mo is the mass of an oxygen isotope 8O17 , MP and Mn are the masses of a proton and a neutron, respectively, the nuclear binding energy of the isotope is

  • ( Mo - 8 MP ) c2 

  • ( Mo - 8 MP - 9 Mc ) c2

  • Mo C2

  • (Mo - 17 Mn ) c2


176.

Hydrogen bomb is based upon

  • fission

  • fusion

  • chemical reaction

  • transmutation


177.

The operation of a nuclear reactor is said to be critical, if the multiplication factor (k) has a value

  • 1

  • 1.5

  • 2.1

  • 2.5


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178.

Half-lives of two radioactive substances A and B are respectively 20 min and 40 min. Initially the samples of A and B have equal number of nuclei. After 80 min the ratio of remaining number of A and B nuclei is

  • 1 : 16

  • 4 : 1

  • 1 : 4

  • 1 : 1


C.

1 : 4

Total number of nuclei remained after n half- lives is 

             N = N12n

Total time given = 80 min

Number of half-lives of A, nA80 min20 min

                                           = 4

Number of half-lives of B, nB80 min40 min

                                           = 2

Number of nuclei remains undecayed

                    N =No 12n

Where No is initial number of nuclei.

             NANB = 12nA12nB

             NANB = 124122               =11614

                NANB = 14


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179.

The half-life of a radio-isotope is 4h. If initial mass of the isotope was 200 g, then mass remaining after 24 h will be

  • 1.042 g

  • 2.084 g

  • 3.125 g

  • 4.167 g


180.

If 82U238 emits 8 α - particles and 6 β - particles, then the resulting nucleus is

  • 82U206

  • 82Pb206

  • 82U210

  • 82U214


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