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OpenStudy (anonymous):

***Simple Nuclear Physics Question***

OpenStudy (anonymous):

OpenStudy (anonymous):

my attempt at #3) No=3.3g X 1mol/69.72g X 6.02e23 nuclei/mol=2.85e22nuclei. so Ro=lambda*No=(.0089hrs-1/3600sec)(2.85e22nuclei)=7.036e16 nuclei/sec(or Bq)

OpenStudy (surry99):

Hint: Remember for first order decay: -dA/dt = k[A] you found k in part 2 and you were given A

OpenStudy (anonymous):

Whats my A?

OpenStudy (anonymous):

isn't A the No im solving for in my equation? @surry99

OpenStudy (anonymous):

nvm I figured it out. My mistake was using the atomic mass for Ga(69.72g/mol), when I was given a Ga-67(66.93g/mol).

OpenStudy (surry99):

good stuff

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