Nuclear Chemistry - Result Question 19

####19. An element ${ } _Z M^{A}$ undergoes an $\alpha$-emission followed by two successive $\beta$-emissions. The element formed is

$(1982,1 M)$

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Answer:

Correct Answer: 19. $2{ } _0 n^{1},{ } _{36} Kr^{82}$

Solution:

  1. ${ } _Z M^{A-4}:{ } _Z M^{A} \longrightarrow{ } _2 He^{4}+2{ } _{-1} e^{0}+{ } _Z M^{A-4}$


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