A nucleus with Z= 92 emits the following in a  sequence :

\alpha, \alpha, \beta ^{-}, \beta ^{-},\alpha, \alpha,\alpha, \alpha,\beta ^{-}, \beta ^{-},\alpha ,\beta ^{+}, \beta ^{+},\alpha. The Z of the resulting nucleus is

  • Option 1)

    76

  • Option 2)

    78

  • Option 3)

    82

  • Option 4)

    74

 

Answers (1)

As we learnt in

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 Minus - decay -

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

- wherein

\bar{\nu }\rightarrow anti \: neutrino

Q\: value =\left [ M_{X}-M_{Y} \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}

 

 

 Total number of \alpha paticle = 8 

Total number of \beta^{-} particle = 4

Total number of \beta^{+} particle = 2

Atomic number = =92-8\times2+4-2=78

Correct option is 2.


Option 1)

76

This is an incorrect option.

Option 2)

78

This is the correct option.

Option 3)

82

This is an incorrect option.

Option 4)

74

This is an incorrect option.

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