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Wave Optics

Question
CBSEENPH12039561

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 4 over denominator straight Z minus 2 end fraction

Solution

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 = fraction numerator straight A minus straight Z minus 4 over denominator straight Z minus 8 end fraction

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