Question and Answers Forum

All Questions      Topic List

Moderm Physics Questions

Previous in All Question      Next in All Question      

Previous in Moderm Physics      Next in Moderm Physics      

Question Number 41361 by Necxx last updated on 06/Aug/18

The half life of radium-226 is 1620 years.Calculate (a)the decay constant (b)the time it takes 60% of a given sample to decay, and (c)the initial activity of 1gm of pure radium-226

$${The}\:{half}\:{life}\:{of}\:{radium}-\mathrm{226}\:{is}\:\mathrm{1620} \\ $$$${years}.{Calculate}\:\left({a}\right){the}\:{decay}\:{constant} \\ $$$$\left({b}\right){the}\:{time}\:{it}\:{takes}\:\mathrm{60\%}\:{of}\:{a}\:{given} \\ $$$${sample}\:{to}\:{decay},\:{and}\:\left({c}\right){the}\:{initial} \\ $$$${activity}\:{of}\:\mathrm{1}{gm}\:{of}\:{pure}\:{radium}-\mathrm{226} \\ $$

Answered by MJS last updated on 06/Aug/18

initial amount = A_0 amount after t years = A_t A_t =A_0 ((1/2))^(t/(1620)) A_t =A_0 e^(−λt) e^(−λt) =((1/2))^(t/(1620)) −λt=(t/(1620))ln (1/2) λ=((ln 2)/(1620))≈4.2787×10^(−4) A_0 =100% 40=100((1/2))^(t/(1620)) (2/5)=((1/2))^(t/(1620)) ln (2/5) =(t/(1620))ln (1/2) t=1620((ln (2/5))/(ln (1/2)))≈2141.5 years

$$\mathrm{initial}\:\mathrm{amount}\:=\:{A}_{\mathrm{0}} \\ $$$$\mathrm{amount}\:\mathrm{after}\:{t}\:\mathrm{years}\:=\:{A}_{{t}} \\ $$$${A}_{{t}} ={A}_{\mathrm{0}} \left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{{t}}{\mathrm{1620}}} \\ $$$${A}_{{t}} ={A}_{\mathrm{0}} \mathrm{e}^{−\lambda{t}} \\ $$$$\mathrm{e}^{−\lambda{t}} =\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{{t}}{\mathrm{1620}}} \\ $$$$−\lambda{t}=\frac{{t}}{\mathrm{1620}}\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\lambda=\frac{\mathrm{ln}\:\mathrm{2}}{\mathrm{1620}}\approx\mathrm{4}.\mathrm{2787}×\mathrm{10}^{−\mathrm{4}} \\ $$$$ \\ $$$${A}_{\mathrm{0}} =\mathrm{100\%} \\ $$$$\mathrm{40}=\mathrm{100}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{{t}}{\mathrm{1620}}} \\ $$$$\frac{\mathrm{2}}{\mathrm{5}}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\frac{{t}}{\mathrm{1620}}} \\ $$$$\mathrm{ln}\:\frac{\mathrm{2}}{\mathrm{5}}\:=\frac{{t}}{\mathrm{1620}}\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${t}=\mathrm{1620}\frac{\mathrm{ln}\:\frac{\mathrm{2}}{\mathrm{5}}}{\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{2}}}\approx\mathrm{2141}.\mathrm{5}\:\mathrm{years} \\ $$

Commented by Necxx last updated on 06/Aug/18

The strength of a radioactive sample may be specified at a given time by its activity and the unit commonly employed is the Curie(Ci) and is defined as: 1Ci=3.70×10^(10) decay/s

$${The}\:{strength}\:{of}\:\:{a}\:{radioactive}\:{sample} \\ $$$${may}\:{be}\:{specified}\:{at}\:{a}\:{given}\:{time}\:{by} \\ $$$${its}\:\boldsymbol{{activity}}\:{and}\:{the}\:{unit}\:{commonly} \\ $$$${employed}\:{is}\:\:{the}\:{Curie}\left({Ci}\right)\:{and}\:{is} \\ $$$${defined}\:{as}: \\ $$$$\mathrm{1}{Ci}=\mathrm{3}.\mathrm{70}×\mathrm{10}^{\mathrm{10}} {decay}/{s} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com