Calcium-45 decays by beta emission. How many beta particles are emitted over a 9

Question

Calcium-45 decays by beta emission. How many beta particles are emitted over a 9.00 minute time period by a sample of^45 Ca that has an activity of 45.0 millcuries? A 90.0 kg worker is exposed to be radiation source above for a period of 9.00 minutes. The kinetic energy of each emitted beta particle is 4.13 Times 10^-14 J. If the worker's body absorbs 19.0 percent of the beta particles emitted over that time period, what is the worker; exposure in millirems? The Relative Biological Effectiveness (RBE) factor for beat particles is 1.

Explanation / Answer

Part A

1 Currie = 3.7*10^10 decays/s

Activity * time = Total no. of emmissions

= 45 *3.7*10^10* 10^-3 * 9 * 60 = Total no. of decays/emmissions

Part B

The total amount of energy = 4.13*10^-14 * No. of decays

Energy absorbed = .19*4.13*10^-14 J * No. of decays

For time period = 9 min

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