An isotope of gallium, 67 Ga, has an atomic number of 31 and a half-life of 78 h

Question

An isotope of gallium, 67Ga, has an atomic number of 31 and a half-life of 78 hours. Consider a small mass of 3.8 grams for 67Ga which is initially pure.

1)Initially, what is the half-life of the gallium?

T1/2o =? hr

2)Initially, what is the initial decay constant of the Ga atom in the sample?

o =? hr-1

3)Initially, what is the decay rate of the gallium?

Ro =? bq

4)What is the half-life of the gallium after 72 hours?

T1/21 =?hr

5)Initially, what is the initial decay constant of the Ga atoms after 72 hours?

1 =? hr-1

6)What is the decay rate of the gallium after 72 hours?

R1 =? bq

Explanation / Answer

Here

a)78 Hours (given)

b) = decay constant

t1/2 = ln(2)/

= ln(2)/t1/2 = ln(2)/78 = 8.8865 * 10-3 hr-1

c) Decay Rate = -dN/dt = N = 8.8865 * 10-3 * (3.8/31) * 6.023 * 1023 = 6.56 * 1020

d) N72 = N0(1-e-8.8865*(10^(-3))*72) = 0.4726N0

0.4726N0/2 = 0.4726N0(1-e-8.8865*(10^(-3))*t)

e-8.8865*(10^(-3))*t = 0.5

8.8865*(10^(-3))*t = ln(0.5)

t = 78 Hours

e) = decay constant = 8.8865 * 10-3 hr-1

f)Decay Rate = N = 8.8865 * 10-3 * 0.4726 * (3.8/31) * 6.023 * 1023 = 3.1 * 1020

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