Mass of iron remains : 3.875 g
Further explanation
Given
62 gram sample of Fe-53 decays for 34.04 minutes.
Required
mass of iron remains
Solution
Hal life of Fe-53 = 8.51 min
[tex]\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}[/tex]
No=62 g
t = 34.04 min
t1/2 = 8.51 min
Input the value :
[tex]\tt Nt=62.\dfrac{1}{2}^{34.04/8.51}\\\\Nt=62.\dfrac{1}{2}^4\\\\Nt=3.875~g[/tex]
