The radioactive isotope phosphorus-32 is often used in biochemicalresearch. Its

Question

The radioactive isotope phosphorus-32 is often used in biochemicalresearch. Its half-life is 14.28 days and it decays by betaemission. (a) Write balanced equation for thedecomposition of P-32. (Use the lowest possible coefficients. Enterthe first (raised) number in the first box, the second (lower)number in the second box, and the symbol of the element in thethird box. Use e to denote an electron particle, n to denote aneutron particle, or p to denote a positron particle. Enter theheaviest particle first and the lightest particle last.) P +

(b) If the original sample were 226 mg ofK332PO4, what amount of P-32remains after 38.8 d?
mg
(a) Write balanced equation for thedecomposition of P-32. (Use the lowest possible coefficients. Enterthe first (raised) number in the first box, the second (lower)number in the second box, and the symbol of the element in thethird box. Use e to denote an electron particle, n to denote aneutron particle, or p to denote a positron particle. Enter theheaviest particle first and the lightest particle last.) P +

(b) If the original sample were 226 mg ofK332PO4, what amount of P-32remains after 38.8 d?
mg
P +

Explanation / Answer

   a) The balenced euation for thedecomposition of p-32.                                                                     3215P  -------->   1432Si   +   -10 b)           The original sample N0   = 226 mg                          Halflife of the reaction t 1/2 = 14.28days                                                           k      =   0.693/ t1/2                                                                  = 0.693/ 14.28                                                              k   = 0.05 day -1                          After 38.8 days the remaing amount N is,                                                                ln (N/N0 ) = -kt                                                                    N/N0      = e-kt                                                                    N = N0 e-kt                                                                        = 226*e- 0.05*38.8                                                                        = 226*0.143                                                                        = 32.32 mg                                                                        = 226*0.143                                                                        = 32.32 mg             after 38.8 days the remaing amount is 32.32mg.

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