Question

If you purchased 0.610 μCi 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.

Answer 1

Answer:

the sample undergo $1.67 \times 10^5$  disintegrations per second.

Explanation:

Given:

The amount of sulphur purchased is 0.671 μCi

To find:

disintegrations per second = ?

Solution:

Some of the conversions are

$1 Ci = 3.8 \times 10^{10} Bq$

1 rad = 0.01 Gy

1Gy = 1 j/kg tissue

1 rem = 0.01 Sv

1Sv = 1 j/Kg

Using these conversions,

The decay rate of this sample is calculated as

$0.671 \mu C i \times \frac{1 \mu C i}{10^{6}} \mu C i \times \frac{0.671 \times 10^{10}}{1 \mu C i}$

= $1.67 \times 10^5$ disintegrations per second