Question

Determine -22 please give the answer​

Answer 1

Step-by-step explanation:

We know 1Ci=3.7×10

10

dps

Activity of Na

22

A=5×10

−3

×(3.7×10

10

) dps

We get A=1.85×10

8

dps

Half life T

1/2

=2.6×365×86400=8.2×10

7

s

From radioactive decay equation, activity of radioactive substance A=

T

1/2

0.693

N

∴ 1.85×10

8

=

8.2×10

7

0.693

N

⟹ N=2.2×10

16

Na

22

atoms

Mass of Na

22

m=

N

A

N

M where M=23 g (atomic weight of Na

22

atom)

∴ m=

6.023×10

23

2.2×10

16

×23

⟹m=8.4×10

−6

g =8.4μg

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