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 In a radioactive decay chain , the initial nucleus is _{90}^{232}\textrm{Th} . At the end there are 6 \alpha - particles and 4 \beta - particles which are emitted. . If the end nucleus is _{Z}^A{}\textrm{X} , A and Z are given by : 

  • Option 1)

    A = 208 ; Z = 80

  • Option 2)

    A = 200 ; Z = 81

  • Option 3)

    A = 202 ; Z = 80

  • Option 4)

    A = 208 ; Z = 82

Answers (1)

best_answer

 

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}

 

Initially _{90}^{232}T

Let finially _{Z}^{A}X

So let reaction

_{90}^{232}Th\rightarrow _{Z}^{A}X+6_{2}^{4}He+4\beta particle

Balance Z - 90 = Z+\left ( 6\times 2 \right )+4\times (-1)=82

Balance A - 232 = A +6\times 4+0=208

So A =208 and Z = 90

 


Option 1)

A = 208 ; Z = 80

Option 2)

A = 200 ; Z = 81

Option 3)

A = 202 ; Z = 80

Option 4)

A = 208 ; Z = 82

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