Question

1. Polonium-209, an α emitter, has a half-life of 102 years.

How many α particles are emitted in 1.5 s from a 1.2 ng sample of 209Po?

2. A 1.0 mg sample of 126Sn decays initially at a rate of 4.6×105 disintegrations/s.

What is the half-life of 126Sn in years?

Answer 1

1) Initial mass of Polonium-209, No 1.2 ng Write the equation Eor decay follows Given half-life, 4a 102 years Calculate the rate constant as follows 0.693 The half life is = 0.693 to 0.693 102 years yearsl k = 6.79x10 Given time, t = 1.5s min 1hr 1 day 1 vear f= 1.5 s> 365 days 60s 24 hr 60 min l= 4,76x10vears Number of a-particl es emitted is n = N-N п- N,- M,exp(-h) Substitute the corresponding values in the above equation. n= 1.2ngx(1-exp(-6.79x103 yearsx4,76x10)) n= 1.2ngx(1-exp(-32.3x 10")) n= 1.2ng x(3.23x10-) 3.9x10ne10gmol 6.023x10 at oms 1ng 209g 1mol n = 0.1124x10 atoms =1124 atoms Therefore, the required solution is 1124 atoms (2) Rate of disintegrati ons, k= 4.6x10 disintegrations/s 0.693 The half life is, ta 0.693 Ev46x10 disintegrati ons/s 1 min 1 day 60 min 24 hr 365 days hr 1 year = 1.5x10 60s tn= 4.76x10 year disintegrati on 126-Sn 1.0 mg Given mass molar mass of 126-Sn 126 g/mol 103g уе ar disintegrations 6.02x10 particles mol x1.0 mg 126 g n= 4.76x10 disintegrati on particles 1mol 1mg 227x101 vear Theref ore, the required soluti on is 2.27x10" year