Menu Close

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-

Tinku Tara 0 Comments
Pin




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

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *