A nucleus of mass M + Δm is at rest and decays into two daughte
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A nucleus of mass M + Δm is at rest and decays into two daughter nuclei of equal mass M/2 each. Speed of light is c.

The speed of daughter nuclei is

  • fraction numerator increment straight m over denominator straight M plus increment straight m end fraction
  • square root of fraction numerator 2 increment straight m over denominator straight M end fraction end root
  • straight c square root of fraction numerator increment straight m over denominator straight M end fraction end root
  • straight c square root of fraction numerator increment straight m over denominator straight M end fraction end root


B.

square root of fraction numerator 2 increment straight m over denominator straight M end fraction end root

Conserving the momentum

Now, from energy conservation and mass -energy equivalence

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A nucleus of mass M + Δm is at rest and decays into two daughter nuclei of equal mass M/2 each. The speed of light is c.
The binding energy per nucleon for the parent nucleus is E1 and that for the daughter nuclei is E2. Then

  • E1 = 2E2

  • E2 = 2E1

  • E1 > E2

  • E1 > E2


D.

E1 > E2

After decay, the daughter nuclei will be more stable hence, binding energy per nucleon will be more than that of their parent nucleus.

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The above is a plot of binding energy per nucleon Eb, against the nuclear mass M; A, B, C, D, E, F correspond to different nuclei. Consider four reactions:

(i) A + B → C + ε  (ii) C → A + B + ε
(iii) D + E → F + ε and (iv) F → D + E +ε
where ε is the energy released? In which reaction is ε positive?

  • (i) and (iv)

  • (i) and (iii)

  • (ii) and (iv)

  • (ii) and (iv)


A.

(i) and (iv)

Binding energy per nucleon of each product is less than that of each reactant.

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A radioactive nucleus (initial mass number A and atomic number Z) emits 3 α–particles and 2 positions. The ratio of number of neutrons to that of protons in the final nucleus will be

  • fraction numerator straight A minus straight Z minus 8 over denominator straight Z minus 4 end fraction
  • fraction numerator straight A minus straight Z minus 4 over denominator straight Z minus 4 end fraction
  • fraction numerator straight A minus straight Z minus 12 over denominator straight Z minus 4 end fraction
  • fraction numerator straight A minus straight Z minus 12 over denominator straight Z minus 4 end fraction

B.

fraction numerator straight A minus straight Z minus 4 over denominator straight Z minus 4 end fraction

In positive beta decay a, proton is transformed into a neutron and a positron is emitted

p+ → n0 + e+
Number of neutrons initially was A-Z
Number of neutrons after decay (A-Z) -3 x 2  (due to alpha particles) + 2 x 1 (due to positive beta decay)
The number of protons will reduce by 8. so, the ratio number of neutrons to that of protons = 

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A radioactive nucleus A with a half-life T, decays into a nucleus B. At t = 0, there is no nucleus B. At some time t, the ratio of the number of B to that of A is 0.3. Then, t is given by

  • t = T log (1.3)

  • straight t space equals space fraction numerator straight T over denominator log space left parenthesis 1.3 right parenthesis end fraction
  • straight t space equals space fraction numerator straight T space log space 2 over denominator 2 space log space 1.3 end fraction
  • straight t space equals space fraction numerator straight T space log space 2 over denominator 2 space log space 1.3 end fraction

D.

straight t space equals space fraction numerator straight T space log space 2 over denominator 2 space log space 1.3 end fraction

At time t

also let initially there are total N0 number of nuclei
NA + NB = N0

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