Can 96-Cm-247 Undergo Alpha Decay?

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The discussion centers on the possibility of 96-Cm-247 undergoing alpha decay. Participants express difficulty in accessing attachments containing calculations. A suggestion is made to present calculations in text format, specifically using LaTeX for clarity. There is an intention to update the discussion with more accessible information. The focus remains on understanding the decay process of 96-Cm-247.
dcarmichael
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Homework Statement
We are asked to calculate the half life of 96-Cm-247 using equations from class, and to compare this calculated value to the real value. When comparing there appears to be a very large discrepancy. I'm wondering if this is to be expected or if there is a calculation error on my part. Ill attach my attempt at a solution
Relevant Equations
attached
nuclear 1.jpg
Nuclear 2.jpg
 
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I am unable to read your attachments. I suggest you type in your calculations. Try to use latex.
 
thank you I will update it.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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