The activity of a radioactive sample is measured as No counts pe

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141.

The activity of a radioactive sample is measured as No counts per minute at t = 0 and No/e counts per minute at t = 5 min. The time (in minute) at which the activity reduces to half its value is

  • loge 2 /5

  • 5/ loge 2

  • 5 log10 2

  • 5 log10 2


D.

5 log10 2

Fraction remains after n half lives

straight N over straight N subscript straight o space equals space open parentheses 1 half close parentheses to the power of straight n space equals space open parentheses 1 half close parentheses to the power of straight t divided by straight T end exponent

Given space space straight N space equals space straight N subscript straight o over straight e space rightwards double arrow space straight N subscript straight o over eN subscript straight o space equals space open parentheses 1 half close parentheses to the power of 5 divided by straight T end exponent
Or space 1 over straight e space equals space open parentheses 1 half close parentheses to the power of 5 divided by straight T end exponent
Taking space log space on space both space side comma space we space get
log space 1 space minus space log space straight e space equals space 5 over straight T space log space 1 half
minus 1 space equals space 5 over straight T space left parenthesis negative log space 2 thin space right parenthesis
straight T space equals space log subscript straight e space 2
Now comma space let space straight t apostrophe space be space the space time space after space which space activity space reduces space to space half
open parentheses 1 half close parentheses space equals open parentheses space 1 half close parentheses to the power of straight t apostrophe divided by 5 space log subscript straight e 2 end exponent
straight t apostrophe space equals space 5 space log subscript straight e space 2

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142.

The decay constant of a ratio isotope is λ. If A1 and A2 are its activities at times t1 and t2 respectively, te number of nuclei which have decayed during the time (t1-t2)

  • A1t1 - A2t2

  • A1 - A2

  • (A1 - A2)/λ

  • (A1 - A2)/λ

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143.

The binding energy per nucleon in deuterium and helium nuclei are 1.1 MeV and 7.0 MeV, respectively. When two deuterium nuclei fuse to form a helium nucleus the energy released in the fusion is

  • 23.6 MeV

  • 2.2 MeV

  • 28.0 MeV

  • 28.0 MeV

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144.

In the nuclear decay given below,

straight X presubscript straight Z presuperscript straight A space rightwards arrow straight Y presubscript straight Z plus 1 end presubscript presuperscript straight A space rightwards arrow space straight B presubscript straight Z minus 1 end presubscript presuperscript straight A minus 4 end presuperscript asterisk times space rightwards arrow space straight B presubscript straight Z minus 1 end presubscript presuperscript straight A minus 4 end presuperscript
the particle emitted in the sequence are

  • β, α, γ

  • γ, β, α, 

  •  β,γ, α

  •  β,γ, α

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145.

The number of beta particles emitted by radioactive substance is twice the number  of alpha particle emitted by it. The resulting daughter is an 

  • isobar of parent

  • isomer of parent

  • isotone of parent

  • isotone of parent

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146.

Two radioactive materials X1 and X2 have decay constant 5 λ respectively. If initially, they have the same number of nuclei, then the ratio of the number of nuclei of X1 to that of X2 will be 1/e after a time

  • λ

  • λ/2

  • 1/4λ

  • 1/4λ

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147.

If M (A, Z), MP and Mn denote the masses of the nucleus straight X presubscript straight Z presuperscript straight A  proton and neutron respectively in units of u (1 u = 931.5 MeV/c2) and BE represents its binding energy  in MeV, then


  • M (A,Z) = ZMp + (A-Z)Mn - BE/c2

  • M (A,Z) = ZMp + (A-Z)Mn + BE

  • M (A,Z) = ZMp + (A-Z)Mn - BF

  • M (A,Z) = ZMp + (A-Z)Mn - BF

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148.

Two nuclei have their mass number in the ratio of 1:3. The ratio of their nuclear densities would be

  • 1:3

  • 3:1

  • (3)1/3 : 1

  • (3)1/3 : 1

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149.

In radioactive decay process, the negatively charged emitted beta 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 produced as a result of collisions between atoms

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150.

A nucleus has straight X presubscript straight Z presuperscript straight A mass represented by  M (A, Z). If Mp and Mn denote the mass of proton and neutron respectively and BE the binding energy (in MeV), then:

  • BE = [M(A,Z)-ZMp - (A-Z)Mn]c2

  • BE = [ZMp + (A-Z)Mn -M(A,Z)]c2

  • BE = [ZMp + AMn - M (A,Z)]c2

  • BE = [ZMp + AMn - M (A,Z)]c2

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