Chemistry - Chemical Kinetics Question with Solution | TestHub

ChemistryChemical KineticsMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasysingle choice

$1 g$ of ${Au}_{79}^{198} (t_{\frac{1}{2}} = 65 h)$ decays by Beta emission and produce stable mercury. How much mercury will be present after 260 h.

Options:

Answer:

Solution:

$A u_{19}^{198} \to H g_{80}^{108} + θ_{- 1}^{0}$

$t_{\frac{1}{2}} = 65 h$

$T = 260 h$

$T = t_{\frac{1}{2}} \times' n'$

Number of Lay live (n) $= \frac{260}{65} = 4$

Amount of undecayed $A u = \frac{N_{0}}{2^{n}} = \frac{1}{2^{4}} = \frac{1}{16} g$

Amount of decayed $Au = 1 - \frac{1}{16} = \frac{15}{16} = 0 .93 g$

Stream:NTA_ABHYASSubject:ChemistryTopic:Chemical KineticsSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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