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

  • Option 1)

    \frac{A-Z-4}{Z-2}

  • Option 2)

    \frac{A-Z-8}{Z-4}

  • Option 3)

    \frac{A-Z-4}{Z-8}

  • Option 4)

    \frac{A-Z-12}{Z-4}

 

Answers (1)

As we learnt

α -decay -

^{A}_{Z}X\rightarrow ^{A-4}_{Z-2}Y+^{4}_{2}He+Q

 

- wherein

Q\: value = \left ( M_{X} -M_{Y}-M_{He}\right )c^{2}

 

 

β plus decay -

^{A}_{Z}X\rightarrow _{Z-1}^{A}Y+\beta ^{+}+ {\nu }+Q \ value      

- wherein

\nu\rightarrow \: neutrino

Q\: value =\left [ M_{X}-M_{Y}-2M_{e} \right ]c^{2}

 

 

 Let's say original nuclei is ^{A}_{Z}X.

^{A}_{Z}X \;\overset{3\alpha}{\rightarrow}\;\ ^{A-12}_{Z-6}Y\; \overset{2\beta^{+}}{\rightarrow}\;^{A-12}_{Z-8}W

Hence,  Number of proton = Z - 8.

Number of neutron = (A - 12) - (Z - 8) = A - Z - 4

Hence, Ratio =\frac{A-Z-4}{Z-8}

Correct option is 3.


Option 1)

\frac{A-Z-4}{Z-2}

This is an incorrect option.

Option 2)

\frac{A-Z-8}{Z-4}

This is an incorrect option.

Option 3)

\frac{A-Z-4}{Z-8}

This is the correct option.

Option 4)

\frac{A-Z-12}{Z-4}

This is an incorrect option.

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