PARTICLES EMITTED \[=x\]and the no. of \[\beta \]-particles emitted =y. Then \[_{92}{{U}^{238}}{{\to }_{82}}P{{B}^{206}}+x_{+2}^{4}\alpha +y\,\,_{-1}^{0}\beta \] equating the mass number on both sides, we get 238 = 206 + 4x + 0 y or 4x = 32, x = 8 equating the atomic number on both sides, we get 92 = 82 + 2x ? y 92 = 82 + 2× 8 ?y y = 6      Hence 8\[\alpha \]and \[6\beta \] are emitted.

"> PARTICLES EMITTED \[=x\]and the no. of \[\beta \]-particles emitted =y. Then \[_{92}{{U}^{238}}{{\to }_{82}}P{{B}^{206}}+x_{+2}^{4}\alpha +y\,\,_{-1}^{0}\beta \] equating the mass number on both sides, we get 238 = 206 + 4x + 0 y or 4x = 32, x = 8 equating the atomic number on both sides, we get 92 = 82 + 2x ? y 92 = 82 + 2× 8 ?y y = 6      Hence 8\[\alpha \]and \[6\beta \] are emitted.

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The number of b-particles emitted in radioactive change \[_{92}{{U}^{238}}{{\to }_{82}}P{{b}^{206}}{{+}_{2}}H{{e}^{4}}\] is       [KCET 2000]

Joint Entrance Exam - Main (JEE Main) Chemistry in Joint Entrance Exam - Main (JEE Main) 11 months ago

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Suppose the no. of a-PARTICLES EMITTED \[=x\]and the no. of \[\beta \]-particles emitted =y. Then \[_{92}{{U}^{238}}{{\to }_{82}}P{{B}^{206}}+x_{+2}^{4}\alpha +y\,\,_{-1}^{0}\beta \] equating the mass number on both sides, we get 238 = 206 + 4x + 0 y or 4x = 32, x = 8 equating the atomic number on both sides, we get 92 = 82 + 2x ? y 92 = 82 + 2× 8 ?y y = 6      Hence 8\[\alpha \]and \[6\beta \] are emitted.

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Posted on 16 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Chemistry in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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