A) \[\mu =\text{1}.\text{33}\]
B) \[\text{1}.\text{5}\times \text{1}{{0}^{\text{8}}}\text{ km}\]
C) \[\text{5}\times \text{1}0{{~}^{14}}\text{Hz}\]
D) \[\text{55}00\text{ { }\!\!\mathrm{\AA}\!\!\text{ }}\]
Correct Answer: C
Solution :
Given \[_{90}T{{h}^{232}}{{\to }_{82}}P{{b}^{208}}\] ??(1) Change in mass number \[=\text{ 232 }-\text{ 2}0\text{8 }=\text{ 24}\] No. of \[\alpha \]-particles emitted \[=\frac{24}{4}=6\] Now equation (1) becomes \[_{90}T{{h}^{232}}{{\xrightarrow{-6\alpha }}_{78}}P{{b}^{208}}{{\xrightarrow{-n\beta }}_{82}}p{{b}^{208}}\] Further change in atomic number is \[82-78=4\] It means atomic no. 78 is increased by 4 to make the atomic no. 82. Therefore 6a-parUcles and 4\[\beta \]-particles will be emitted..
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