Nuclei

Sponsor Area

Question
CBSEENPH12039847

A mixture consists of two radioactive materials A1 and A2 with half-lives of 20 s and 10 s respectively. Initially, the mixture has 40 g of A1 and 160 g of A2. The amount of the two in the mixture will become equal after

  • 60 s

  • 80 s

  • 20 s

  • 40 s

Solution

D.

40 s

For 40 g amount
WiredFaculty
So, after 40 A1 and A2 remains same.

Sponsor Area

Question
CBSEENPH12039925

A nucleus X presubscript straight n presuperscript straight m emits one α - particle and two β- particles. The resulting nucleus is 

  • straight Z presubscript straight n presuperscript straight m minus 6 end presuperscript
  • straight X presubscript straight n presuperscript straight m minus 4 end presuperscript space space
  • straight Y presubscript straight n minus 2 end presubscript presuperscript straight m minus 4 end presuperscript
  • straight Z presubscript straight n minus 4 end presubscript presuperscript straight m minus 6 end presuperscript

Solution

B.

straight X presubscript straight n presuperscript straight m minus 4 end presuperscript space space straight X presubscript straight n presuperscript straight m space rightwards arrow with straight alpha on top space straight X presubscript straight n minus 2 end presubscript superscript straight m minus 4 end superscript space rightwards arrow with 2 straight beta on top space straight X presubscript straight n superscript straight m minus 4 end superscript

Question
CBSEENPH12040069

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 = M (A,Z)- ZMp - (A-Z) Mn

Solution

B.

BE = [ZMp + (A-Z)Mn -M(A,Z)]c2

In the case of formation of a nucleus, the evolution of energy equal to the binding energy of the nucleus takes place due to the disappearance of a fraction of the total mass. If the quantity of mass disappearing is ΔM, then the binding energy is
BE = ΔMc2
From the above discussion, it is clear that the mass of the nucleus must be less than the sum of the masses of the constituent neutrons and protons. We can then write.


ΔM= [ZMp + (A-Z)Mn -M(A,Z)
Where M (A, Z)  is the mass of the atom of mass number A and atomic number Z. Hence, the binding energy of the nucleus is
BE = [ZMp + (A-Z)Mn -M(A,Z)]c2

Where N = A -Z = number of neutrons.

Question
CBSEENPH12039948

A nucleus of uranium decays at rest into nuclei of thorium and helium. Then, 

  • the helium nucleus has more kinetic energy than the thorium nucleus

  • the helium nucleus has less momentum than the thorium nucleus

  • the helium nucleus has more momentum than the thorium nucleus

  • the helium nucleus has less kinetic energy than the thorium nucleus 

Solution

A.

the helium nucleus has more kinetic energy than the thorium nucleus

WiredFaculty

Question
CBSEENPH12039886

A radio isotope X with a half life 1.4 x 109 yr decays of Y which is stable. A sample of the rock from a cave was found to contain X and Y in the ratio 1:7. The age of the rock is,

  • 1.96 x 109 yr

  • 3.92 x 109 yr

  • 4.20 x 109 yr

  • 8.40 x 109 yr

Solution

C.

4.20 x 109 yr

Ratio of X:Y is given = 1:7
That is,
straight m subscript straight x over straight m subscript straight y equals 1 over 7
rightwards double arrow space 7 straight m subscript straight x space equals space straight m subscript straight y
Let, the initial total mass is m.

WiredFaculty
Therefore, time taken to become 1/8 unstable part
= 3 x T1/2
= 3 x 1.4 x 109
= 4.2 x 109 y

82