Part A If you purchased 3.51 ? Ci of sulfur-35, how many disintegrations per sec

Question

Part A

If you purchased 3.51?Ci of sulfur-35, how many disintegrations per second does the sample undergo when it is brand new?

Express your answer numerically in disintegrations per second.

Part B

A 85.0-kg person is exposed to 369rad of radiation. How many joules did this person absorb?

Measuring Radioactivity Radiation intensity is expressed in different ways, depending on how it is being measured. Mechanical measurements use becquerels (Bq) or curies (Ci). Energy absorbed by tissue is expressed in grays (Gy) or rads (rad). Tissue damage is expressed in sieverts (Sv) or rem. These units are related as follows: 1 Bq = 1 disintegration / s (dps) 1 Ci = 3.7 times 1010 Bq 1 rad = 0.01 Gy, 1 Gy = 1 J / (kg tissue) 1 rem = 0.01 Sv, 1 Sv = 1 J/kg If you purchased 3.51 mu Ci of sulfur - 35, how many disintegrations per second does the sample undergo when it is brand new? Express your answer numerically in disintegrations per second. A 85.0-kg person is exposed to 369rad of radiation. How many joules did this person absorb? Express your answer numerically in joules.

Explanation / Answer

PART A

1Ci = 3.7 x 1010 Bq

3.51 ? Ci = 3.51 x 10-6 x 3.7 x 1010 Bq

                  = 12.98 x 104 Bq

1 Bq = 1 dps

12.98 x 104 Bq = 12.98 x 104 dps

Number of disintegrations per second = 12.98 x 104 dps

PART B

1 rad = 0.01 Gy

100 rad = 1 Gy

369 rad = 1 x 369/100 = 3.69 Gy

1 Gy = 1 J/kg tissue

3.69 Gy = 3.69 J/kg tissue

Radiation absorbed for 1 kg = 3.69 J

Radiation absorbedby85 kg person = 85 x 3.69 J

                                                         = 313.65 J

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