1.

.^(64)Cu (hlaf lifre = 12.8h)deacays by beta^(-) emission (38%), beta^(+)- emission (19%) and electron capute (43%). Write the decay products and calculatepartial half-live for each of the decay processes.

Answer» <html><body><p></p>Solution :We know,<br/> <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/DPB_PHY_CHM_IX_C04_E01_051_S01.png" width="80%"/> <br/> Given `lambda_(av) = (0.693)/(12.8) hr^(-1)`<br/>`:. lambda_(1) +lambda_(2) +lambda_(3) =lambda_(av) = (0.693)/(12.8) = 5.41xx10^(-2) hr^(-1)` .....(1) <br/> <a href="https://interviewquestions.tuteehub.com/tag/also-373387" style="font-weight:bold;" target="_blank" title="Click to know more about ALSO">ALSO</a> for parallel <a href="https://interviewquestions.tuteehub.com/tag/path-11833" style="font-weight:bold;" target="_blank" title="Click to know more about PATH">PATH</a> <a href="https://interviewquestions.tuteehub.com/tag/decay-945588" style="font-weight:bold;" target="_blank" title="Click to know more about DECAY">DECAY</a> <br/> `lambda_(1)` = Fractionlal yield of `._(30)^(64)Zn xx lambda_(av)` <br/> `lambda_(2)` = Fractionalyeild of `._(28)^(64) Ni xx lambda_(av)`<br/> `lambda_(3)` = Fractionlal yield of `._(28)^(64) Ni xx lambda_(av)` <br/> `:. (lambda_(1))/(lambda_(2)) = (38)/(19)` ....(2) <br/> `(lambda_(1))/(lambda_(3)) = (38)/(43)` ....(3) <br/> From eqs. (1), (2) and (3) <br/> `lambda_(1) = 2.056xx10^(-2) hr^(-1)` <br/> `lambda_(2) = 1.028xx10^(-2) hr^(-1)` <br/> `lambda_(3) = 2.327xx10^(-2) hr^(-1)` <br/> `:. t_(1//2)` for `beta^(-)` emission <br/> `= (0.693)/(2.056xx10^(-2)) = 33.70 hr` <br/>`t_(1//2)` for `beta^(+)` emission <br/> `= (0.693)/(1.028xx10^(-2)) = 67.41 hr` <br/>`t_(1//2)` for electron capture <br/> `= (0.693)/(2.327xx10^(-2)) = 29.28 hr`</body></html>


Discussion

No Comment Found

Related InterviewSolutions