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  • Need clarity, kindly explain! - Chemical kinetics - JEE Main-2

Consider the following nuclear reactions:

_{92}^{238}\textrm{M}\rightarrow _{Y}^{X}\textrm{N}+2\; _{2}^{4}\textrm{He}\; \; ;  _{Y}^{X}\textrm{N}\rightarrow\; _{B}^{A}\textrm{L}+2\; \beta ^{+} The number of neutrons in the element L is

  • Option 1)

    142

  • Option 2)

    144

  • Option 3)

    140

  • Option 4)

    146

 

Answers (1)

best_answer

As we learned

\beta ^{+}- - decay -

In this type of decay, a proton first break down into neutrons and positron. Then the positron is emitted from the nucleus, while the neutron stay inside the nucleus. The resulting nucleus has one less proton and one more neutron.  

- wherein

_{Z}^{A}x\rightarrow _{Z-1}^{A}y+e^{+}

 

 Nuclear Reaction

_{92}^{238}M\rightarrow _{88}^{230}N+2\: _{2}^{H}He

_{88}^{230}N\rightarrow _{86}^{230}L+2\beta ^{+}

\Rightarrow no\: o\! f \: neutron \: in \: the\: element\: L= 230-86=144


Option 1)

142

Option 2)

144

Option 3)

140

Option 4)

146

Posted by

Aadil

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