{ }^{234}Th_{90} gets converted to { }^{206}P | vedclass

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(D) The radioactive decay process is represented as: ${ }^{234}Th_{90} \longrightarrow { }^{206}Pb_{82} + x({ }^{4}He_{2}) + y({ }^{0}e_{-1})$.$1$. Fo

Vedclass · Accepted Answer

$7$ and $6$

Vedclass · Answer

(D) The radioactive decay process is represented as: ${ }^{234}Th_{90} \longrightarrow { }^{206}Pb_{82} + x({ }^{4}He_{2}) + y({ }^{0}e_{-1})$.$1$. For mass number conservation:$234 = 206 + 4x$$4x = 234 - 206 = 28$$x = 7$ (Number of alpha particles).$2$. For atomic number conservation:$90 = 82 + 2x - y$$90 = 82 + 2(7) - y$$90 = 82 + 14 - y$$90 = 96 - y$$y = 6$ (Number of beta particles).Therefore,the number of alpha and beta particles lost are $7$ and $6$ respectively.

${ }^{234}Th_{90}$ gets converted to ${ }^{206}Pb_{82}$ through a series of radioactive decay processes. The number of alpha and beta particles lost in this transformation respectively,are

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